The limit of cosine of $3$ times $x$ minus cosine of $4$ times $x$ divided by $x$ times sine of $2$ times $x$ is indeterminate as the value of $x$ tends to zero, as per the limit by direct substitution.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cos{3x}-\cos{4x}}{x\sin{2x}}}$ $\,=\,$ $\dfrac{0}{0}$
It is proved that finding the limit by direct substitution is not useful for this function. So, let’s analyze the rational trigonometric function for preparing a plan of finding the limit.
According to our analysis, we cannot do anything in denominator to avoid the zero. So, let’s try to do something to the expression in the numerator.
Now, let’s focus on the expression in the numerator. There are two cosine functions with multiple angles and they both are connected by a minus sign but the cos of $4x$ cannot be subtracted from cos of $3x$ mathematically due to unknown value of $x$. However, the difference between cosine of multiple angle functions can be written in product form of sine functions.
According to difference to product identity of cosine functions,
$\cos{a}$ $-$ $\cos{b}$ $\,=\,$ $-2\sin{\Big(\dfrac{a+b}{2}\Big)}\sin{\Big(\dfrac{a-b}{2}\Big)}$
Let’s take $a = 3x$ and $b = 4x$. Now, convert the expression in the numerator by the above trigonometric identity.
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cos{3x}-\cos{4x}}{x\sin{2x}}}$ $\,=\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-2\sin{\Big(\dfrac{3x+4x}{2}\Big)}\sin{\Big(\dfrac{3x-4x}{2}\Big)}}{x\sin{2x}}}$
Now, let’s simplify the trigonometric expression in the numerator of the rational function.
$\,=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-2\sin{\Big(\dfrac{7x}{2}\Big)}\sin{\Big(\dfrac{-x}{2}\Big)}}{x\sin{2x}}}$
Observe the factors in both numerator and denominator, the majority of the factors are in terms of sine function and it is a good sign to find the limit by using the trigonometric limit rule in sine function.
Now, let’s focus on simplifying trigonometric expression further.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-2\sin{\Big(\dfrac{7x}{2}\Big)}\sin{\Big(-\dfrac{x}{2}\Big)}}{x\sin{2x}}}$
In numerator, there is a sine function with negative angle. It can be simplified by using even or odd identity of sine function.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-2\sin{\Big(\dfrac{7x}{2}\Big)}\bigg(-\sin{\Big(\dfrac{x}{2}\Big)}\bigg)}{x\sin{2x}}}$
Now, let’s begin the process of simplifying the numerator to possible level.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{-2\sin{\Big(\dfrac{7x}{2}\Big)}\bigg((-1) \times \sin{\Big(\dfrac{x}{2}\Big)}\bigg)}{x\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{(-2) \times \sin{\Big(\dfrac{7x}{2}\Big)} \times (-1) \times \sin{\Big(\dfrac{x}{2}\Big)}}{x\sin{2x}}}$
Use the commutative property to change the positions of the factors in the expression.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{(-2) \times (-1) \times \sin{\Big(\dfrac{7x}{2}\Big)} \times \sin{\Big(\dfrac{x}{2}\Big)}}{x\sin{2x}}}$
Now, multiply the integers $-2$ and $-1$ to find their product arithmetically.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)} \times \sin{\Big(\dfrac{x}{2}\Big)}}{x\sin{2x}}}$
Now, it is time to find the limit of the rational trigonometric function by a trigonometric limit rule.
According to trigonometric limit rule in sine function, a sine function should have its denominator same as its inside angle. So, let’s try to make each sine function consists the denominator same as its inside angle.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)} \times \sin{\Big(\dfrac{x}{2}\Big)}}{x \times \sin{2x}}}$
Let’s focus on first sine function and it has an angle $7x$ divided by $2$ in it. So, this sine function should have $7x$ divided by $2$ as its denominator, whereas $x$ is already there in denominator. Now, let’s split the rational function as a product of two rational functions as per the multiplication of fractions.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{x} \times \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}\Bigg)}$
The limit of product of two functions can be evaluated by the product of their limits as per the product rule of limits.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
Now, look at the first factor and let’s try to find its limit. $x$ is already there in denominator but $x$ should have $7$ as its coefficient and $2$ as divisor of their product. So, let’s try to make the denominator exactly same as the angle inside the sine function.
According to a property of factors, one is a factor of every quantity. So, $x$ can be written as a product of one and $x$ in denominator.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{1 \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
The quotient of $2$ divided by $2$ is equal to one. So, $1$ can be written as $2$ divided by $2$ because the variable $x$ should have $2$ as its divisor in this case.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{2}{2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
The number $2$ can be written in factors form as per the factors of two.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{2 \times 1}{1 \times 2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
Now, split the fraction as a product of two fractions as per multiplication of fractions.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{2}{1} \times \dfrac{1}{2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
Now, divide the $2$ by $1$ and the quotient of $2$ divided by $1$ is equal to $2$.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{2 \times \dfrac{1}{2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cancel{2} \times \sin{\Big(\dfrac{7x}{2}\Big)}}{\cancel{2} \times \dfrac{1}{2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{1 \times \dfrac{1}{2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
Now, let’s express the variable $x$ in fraction form to multiply it by a fraction in the denominator.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{1}{2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{1}{2} \times \dfrac{x}{1}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{1 \times x}{2 \times 1}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
The expression in the denominator should have $7$ as coefficient. So, let’s repeat the same procedure obtain the denominator same as the angle inside the sine function.
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(1 \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{7}{7} \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{7 \times 1}{1 \times 7} \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{7}{1} \times \dfrac{1}{7} \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(7 \times \dfrac{1}{7} \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(7 \times \dfrac{1 \times \sin{\Big(\dfrac{7x}{2}\Big)}}{7 \times \dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(7 \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{7 \times \dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(7 \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{7}{1} \times \dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(7 \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{7 \times x}{1 \times 2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(7 \times \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{7x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
The number $7$ is a factor and it is a constant. So, it can be excluded from the limiting operation by the constant multiple limits rule.
$=\,\,$ $7 \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{7x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
The second factor is almost same as the trigonometric limit rule but the input value should also be same as the denominator. So, let’s set the input of the limiting operation.
$(1).\,\,$ If $x \,\to\, 0$, then $7 \times x \,\to\, 7 \times 0$. Therefore, $7x \,\to\, 0$
$(2).\,\,$ If $7x \,\to\, 0$, then $7x \div 2 \,\to\, 0 \div 2$. Therefore, $\dfrac{7x}{2} \,\to\, 0$
The two above two steps proved that if $x$ approaches to zero, then $7$ times $x$ divided by $2$ also approaches to zero.
$=\,\,$ $7 \times \displaystyle \large \lim_{{\Large \frac{7x}{2}} \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{7x}{2}\Big)}}{\dfrac{7x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
According to trigonometric limit rule in sine function, the limit of sine of $7$ times $x$ divided by $2$, divided by $7$ times $x$ divided by $2$ as the value of $7$ times $x$ divided by $2$ is equal to one.
$=\,\,$ $7 \times 1$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}}$
It is time to find the limit of the remaining function. There is a sine function in both numerator and denominator. If each sine function has denominator exactly same as the angle inside the sine function, then the trigonometric limit rule in sine function can be used to find the limit.
The sine of $x$ divided by $2$ should have $x$ divided by $2$ as its denominator. Similarly, the sine of double angle $2x$ should also have $2x$ as its denominator. So, both sine functions should have $x$ in their denominators. Now, let’s try to include $x$ in the function.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(1 \times \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}\Bigg)}$
The quotient of $x$ divided by $x$ is equal to one. So, the number $1$ can be written as the quotient of $x$ divided by $x$.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{x}{x} \times \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\sin{2x}}\Bigg)}$
We know, $1$ is a factor of every quantity. So, the variable $x$ can be written as a product of $1$ and $x$. Similarly, write each sine function as a product of one and sine function.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{x \times 1}{1 \times x} \times \dfrac{\sin{\Big(\dfrac{x}{2}\Big) \times 1}}{1 \times \sin{2x}}\Bigg)}$
Now, split both rational functions as a product of two rational functions as per the multiplication rule of fractions.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{x}{1} \times \dfrac{1}{x} \times \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{1} \times \dfrac{1}{\sin{2x}}\Bigg)}$
Use the commutative property to change the positions of the factors in the multiplication.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{x} \times \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{1} \times \dfrac{x}{1} \times \dfrac{1}{\sin{2x}}\Bigg)}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{x} \times \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{1} \times x \times \dfrac{1}{\sin{2x}}\Bigg)}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{1 \times \sin{\Big(\dfrac{x}{2}\Big)}}{x \times 1} \times x \times \dfrac{1}{\sin{2x}}\Bigg)}$
Now, multiply the both factors in numerator and denominator to find their products.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{x} \times x \times \dfrac{1}{\sin{2x}}\Bigg)}$
The sine of double angle is in denominator. So, the variable $x$ should be its denominator and it is possible to express $x$ as the reciprocal of reciprocal of $x$.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{x} \times \dfrac{1}{\Big(\dfrac{1}{x}\Big)} \times \dfrac{1}{\sin{2x}}\Bigg)}$
Now, use the multiplication of fractions to get the product of fractions.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{x} \times \dfrac{1 \times 1}{\Big(\dfrac{1}{x}\Big) \times \sin{2x}} \Bigg)}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{x} \times \dfrac{1}{\dfrac{1}{x} \times \dfrac{\sin{2x}}{1}} \Bigg)}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{x} \times \dfrac{1}{\dfrac{1 \times \sin{2x}}{x \times 1}}\Bigg)}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{x} \times \dfrac{1}{\dfrac{\sin{2x}}{x}}\Bigg)}$
Let’s use the product rule of limits to find the limit of product of two functions by finding the product of their limits.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
The angle inside the sine function is $x$ divided by $2$. So, the denominator should also be same. Let’s use the same procedure to obtain the number $2$ as the denominator of $x$.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{1 \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{2}{2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{2 \times 1}{1 \times 2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{2}{1} \times \dfrac{1}{2} \times x}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{2 \times \dfrac{1}{2} \times \dfrac{x}{1}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{2 \times \dfrac{1 \times x}{2 \times 1}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{2 \times \dfrac{x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
Now, the factor $2$ in the denominator can be excluded from the limiting operation by the constant multiple limits formula.
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1 \times \sin{\Big(\dfrac{x}{2}\Big)}}{2 \times \dfrac{x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $7$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{2} \times \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{x}{2}}\Bigg)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $7$ $\times$ $\dfrac{1}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
Now, multiply the constant factors to find the product of the fractions.
$=\,\,$ $\dfrac{7}{1}$ $\times$ $\dfrac{1}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $\dfrac{7 \times 1}{1 \times 2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
Now, let’s focus on the second factor to find the limit of sine of $x$ divided by $2$ divided by $x$ divided by $2$ as the value of $x$ approaches to zero.
If $x \,\to\, 0$, then $x \div 2 \,\to\, 0 \div 2$. Therefore, $\dfrac{x}{2} \,\to\, 0$
It proved that when $x$ approaches to zero, the quotient of $x$ divided by $2$ also tends to zero.
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{{\Large \frac{x}{2}} \,\to\, 0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\dfrac{x}{2}}}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
According to trigonometric limit rule in sine function, the limit of sine of $x$ divided by $2$, divided by $x$ divided by $2$ as the value of $x$ divided by $2$ approaches zero is equal to one.
$=\,\,$ $\dfrac{7}{2}$ $\times$ $1$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
Let’s use the multiplication of the fractions to find the product of the fractions.
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\dfrac{1}{1}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $\dfrac{7 \times 1}{2 \times 1}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{x}}}$
Finally, let’s find the limit of the remaining reciprocal function to find the limit of the given trigonometric rational function.
The angle inside the sine function is $2$ times $x$, whereas the denominator requires the number $2$ as a coefficient of $x$. So, let’s try to include the number $2$ as a coefficient of $x$ by writing rational function in denominator with factor one.
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{1 \times \dfrac{\sin{2x}}{x}}}$
The variable $x$ requires $2$ as its coefficient. So, write the one as the quotient of $2$ divided by $2$ in the denominator and let’s use the multiplication of fractions to complete our task.
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{2}{2} \times \dfrac{\sin{2x}}{x}}}$
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{2 \times 1}{1 \times 2} \times \dfrac{\sin{2x}}{x}}}$
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{2}{1} \times \dfrac{1}{2} \times \dfrac{\sin{2x}}{x}}}$
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{2 \times \dfrac{1}{2} \times \dfrac{\sin{2x}}{x}}}$
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{2 \times \dfrac{1 \times \sin{2x}}{2 \times x}}}$
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{2 \times \dfrac{\sin{2x}}{2x}}}$
The angle inside the sine function is exactly appearing in the denominator. It means, the function is in the form of trigonometric limit rule in the form of sine function. Now, let’s try to eliminate the constant factors from the rational function by the multiplication of fractions.
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1 \times 1}{2 \times \dfrac{\sin{2x}}{2x}}}$
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \Bigg(\dfrac{1}{2} \times \dfrac{1}{\dfrac{\sin{2x}}{2x}}\Bigg)}$
Finally, use the constant multiple limit rule to eliminate the constant factor from the limits operation.
$=\,\,$ $\dfrac{7}{2}$ $\times$ $\dfrac{1}{2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{2x}}}$
Now, multiply the fractions to find the product of them.
$=\,\,$ $\dfrac{7 \times 1}{2 \times 2}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{2x}}}$
$=\,\,$ $\dfrac{7}{4}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{1}{\dfrac{\sin{2x}}{2x}}}$
The limit of reciprocal of a rational function can be evaluated by finding the reciprocal of its limit as per the reciprocal rule of limits.
$=\,\,$ $\dfrac{7}{4}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\sin{2x}}{2x}}}$
If $x \,\to\, 0$, then $2 \times x \,\to\, 2 \times 0$, therefore $2x \,\to\, 0$.
It proved that when the value of $x$ approaches to zero, the value of two times $x$ also tends to zero. Therefore, the input of limit operation can be modified mathematically to obtain the function in required form.
$=\,\,$ $\dfrac{7}{4}$ $\times$ $\dfrac{1}{\displaystyle \large \lim_{2x \,\to\, 0}{\normalsize \dfrac{\sin{2x}}{2x}}}$
According to trigonometric limit rule, the limit of sine of two times $x$ divided by two times $x$ as the value of two times $x$ approaches to zero is equal to one.
$=\,\,$ $\dfrac{7}{4}$ $\times$ $\dfrac{1}{1}$
Now, let’s multiply the fractions to find their product and it will be the limit of the given trigonometric rational function.
$=\,\,$ $\dfrac{7 \times 1}{4 \times 1}$
$=\,\,$ $\dfrac{7}{4}$
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