Evaluate $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\ln{(\cos{x})}}{\sqrt[4]{1+x^2}-1}}$

In this limit problem, a rational expression is defined in terms of logarithmic, trigonometric and algebraic form by considering $x$ as a variable.

$\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\ln{(\cos{x})}}{\sqrt[\Large 4]{1+x^2}-1}}$

The natural logarithm of cosine of angle $x$ is a mathematical expression in the numerator. The subtraction of one from the fourth root of one plus $x$ squared is a mathematical expression in the denominator. The quotient of them is defined as a rational expression and we have to evaluate this mathematical expression as $x$ closer to $0$.

Evaluate the Limit by Direct substitution

Substitute $x$ is equal to zero in the rational function to find the limit of the given rational expression as the variable $x$ approaches zero.

$=\,\,\,$ $\dfrac{\ln{(\cos{0})}}{\sqrt[\Large 4]{1+(0)^2}-1}$

According to the trigonometry, the cosine of zero radian is equal to one.

$=\,\,\,$ $\dfrac{\ln{(1)}}{\sqrt[\Large 4]{1+0^2}-1}$

$=\,\,\,$ $\dfrac{\ln{(1)}}{\sqrt[\Large 4]{1+0}-1}$

$=\,\,\,$ $\dfrac{\ln{(1)}}{\sqrt[\Large 4]{1}-1}$

According to arithmetic mathematics, the fourth root of one is also one.

$=\,\,\,$ $\dfrac{\ln{(1)}}{1-1}$

The natural logarithm of one is equal to zero as per the log of one rule.

$=\,\,\,$ $\dfrac{0}{0}$

Therefore, we have evaluated that the limit of the given rational expression is indeterminate as $x$ tends to zero. It clears that it is not recommendable to use the direct substitution method and it signals us to use an alternative method for calculating the limit of the given rational expression.

Convert the function into Logarithmic limit rule

In this limit problem, there is a natural logarithmic function in the numerator position of the rational expression and we have a limit rule in logarithmic function. Hence, let’s try to transform the rational expression in the form of the logarithm limit rule.

Firstly, add one and subtract one from the cosine of angle $x$ inside the natural logarithmic function.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\ln{(\cos{x}+1-1)}}{\sqrt[\Large 4]{1+x^2}-1}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\ln{(1+\cos{x}-1)}}{\sqrt[\Large 4]{1+x^2}-1}}$

For using the natural logarithmic limit rule in this case, the natural logarithmic function should contain an expression $\cos{x}-1$ as its denominator. So, let’s try to insert it in the rational expression acceptably.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{\ln{(1+\cos{x}-1)}}{\sqrt[\Large 4]{1+x^2}-1}}$ $\times$ $1 \Bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{\ln{(1+\cos{x}-1)}}{\sqrt[\Large 4]{1+x^2}-1}}$ $\times$ $\dfrac{\cos{x}-1}{\cos{x}-1} \Bigg)$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{\ln{(1+\cos{x}-1)}}{\cos{x}-1}}$ $\times$ $\dfrac{\cos{x}-1}{\sqrt[\Large 4]{1+x^2}-1} \Bigg)$

The limit of the product of the functions can be factored as the product of their limits by the product rule of limits.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\ln{(1+\cos{x}-1)}}{\cos{x}-1}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\cos{x}-1}{\sqrt[\Large 4]{1+x^2}-1}}$

In the first factor of the expression, the rational expression is the same as the rational expression in the logarithmic limit rule but the input of the limit operation should also be the same. So, let’s set the input of the limit operation same as the log limit formula.

$(1).\,\,\,$ If $x \,\to\,0$ then $\cos{x} \,\to\, \cos{(0)}$. Therefore, $\cos{x} \,\to\, 1$

$(2).\,\,\,$ If $\cos{x} \,\to\, 1$ then $\cos{x}-1 \,\to\, 1-1$. Therefore, $\cos{x}-1 \,\to\, 0$

The two steps have cleared that the value of $\cos{x}-1$ approaches $0$ when $x$ closer to $0$.

$=\,\,\,$ $\displaystyle \large \lim_{\cos{x}-1\,\to\,0}{\normalsize \dfrac{\ln{(1+\cos{x}-1)}}{\cos{x}-1}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\cos{x}-1}{\sqrt[\Large 4]{1+x^2}-1}}$

For your convenience, assume $y \,=\, \cos{x}-1$ but keep the second factor in terms of $x$.

$=\,\,\,$ $\displaystyle \large \lim_{y\,\to\,0}{\normalsize \dfrac{\ln{(1+y)}}{y}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\cos{x}-1}{\sqrt[\Large 4]{1+x^2}-1}}$

As per the logarithmic limit rule, the quotient of logarithm of $1$ plus $y$ by $y$ as $y$ approaches $0$ is equal to one.

$=\,\,\,$ $1$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\cos{x}-1}{\sqrt[\Large 4]{1+x^2}-1}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\cos{x}-1}{\sqrt[\Large 4]{1+x^2}-1}}$

Convert the function into Trigonometric limit rule

The expression in the numerator of the simplified rational expression is $\cos{x}-1$, which is a trigonometric expression. In limits, we have two trigonometric limit rules in which one limit rule is in terms of sine function. According to the trigonometric identities, the trigonometric expression $\cos{x}-1$ can be transformed into sine function. So, let’s convert it in terms of sine function by cosine half angle identity.

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{-(1-\cos{x})}{\sqrt[\Large 4]{1+x^2}-1}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{-\Bigg(2\sin^2{\Big(\dfrac{x}{2}\Big)}\Bigg)}{\sqrt[\Large 4]{1+x^2}-1}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{-2\sin^2{\Big(\dfrac{x}{2}\Big)}}{\sqrt[\Large 4]{1+x^2}-1}}$

$=\,\,\,$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(-2 \times \dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

The factor $-2$ is a constant and it can be taken out from the limiting operation by the constant multiple rule of limits.

$=\,\,\,$ $-2 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\sqrt[\Large 4]{1+x^2}-1}}$

Now, we have to transform the rational expression into the trigonometric limit rule of sine function. The sine function in the numerator is in square form. So, it is required that this function should have the square form angle inside the sine function as its denominator. So, let us take the necessary steps to obtain the required form.

$=\,\,\,$ $-2 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\sqrt[\Large 4]{1+x^2}-1}}$ $\times$ $1\Bigg)$

$=\,\,\,$ $-2 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\sqrt[\Large 4]{1+x^2}-1}}$ $\times$ $\dfrac{\Big(\dfrac{x}{2}\Big)^2}{\Big(\dfrac{x}{2}\Big)^2}\Bigg)$

$=\,\,\,$ $-2 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)^2}}$ $\times$ $\dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)$

Use the product rule of limits for expanding the limiting operation for the product of the functions.

$=\,\,\,$ $-2 \times \Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin^2{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)^2}}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

Concentrate on the trigonometric rational expression, the quotient of the squares can be written as square of their quotient by the power of a quotient rule.

$=\,\,\,$ $-2 \times \Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \Bigg(\dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)}\Bigg)^2}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

We can use the power rule of limits for preparing the trigonometric function.

$=\,\,\,$ $-2 \times \Bigg(\Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)}\Bigg)^2}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

If $x\,\to\,0$, then $\dfrac{x}{2}\,\to\,\dfrac{0}{2}$. Therefore, $\dfrac{x}{2}\,\to\,0$

$=\,\,\,$ $-2 \times \Bigg(\Bigg(\displaystyle \large \lim_{\Large \frac{x}{2}\,\large\to\,0}{\normalsize \dfrac{\sin{\Big(\dfrac{x}{2}\Big)}}{\Big(\dfrac{x}{2}\Big)}\Bigg)^2}$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

According to the trigonometric limit rule in sine function, the limit of the rational expression as $\dfrac{x}{2}$ approaches zero is equal to one.

$=\,\,\,$ $-2 \times \Bigg((1)^2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

$=\,\,\,$ $-2 \times \Bigg(1^2$ $\times$ $\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

$=\,\,\,$ $-2 \times \Bigg(1 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

$=\,\,\,$ $-2 \times \Bigg(\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}\Bigg)}$

$=\,\,\,$ $-2 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\Big(\dfrac{x}{2}\Big)^2}{\sqrt[\Large 4]{1+x^2}-1}}$

Convert the function into Trigonometric limit rule

The rational expression can be written in multiplicative inverse form.

$=\,\,\,$ $-2 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{1}{\dfrac{\sqrt[\Large 4]{1+x^2}-1}{\Big(\dfrac{x}{2}\Big)^2}}}$

Use the reciprocal rule of the limits for evaluating the limit of the multiplicative inverse of the rational expression.

$=\,\,\,$ $-2 \times \dfrac{1}{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt[\Large 4]{1+x^2}-1}{\Big(\dfrac{x}{2}\Big)^2}}}$

We can use the power rule of a quotient to expand it as a quotient of their squares.

$=\,\,\,$ $-2 \times \dfrac{1}{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt[\Large 4]{1+x^2}-1}{\dfrac{x^2}{2^2}}}}$

Now, we have to focus on simplifying this mathematical expression.

$=\,\,\,$ $-2 \times \dfrac{1}{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt[\Large 4]{1+x^2}-1}{\dfrac{x^2}{4}}}}$

$=\,\,\,$ $-2 \times \dfrac{1}{\displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{4 \times \Big(\sqrt[\Large 4]{1+x^2}-1\Big)}{x^2}}}$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt[\Large 4]{1+x^2}-1}{x^2}}}$

The rational expression in the limit operation is similar to the quotient difference rule limit rule in exponential form. So, let’s take some mathematical steps to obtain it in the required form.

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt[\Large 4]{1+x^2}-\sqrt[\Large 4]{1}}{x^2}}}$

Now, add a plus one to the denominator and also subtract $1$ from it for transforming the rational expression in required form.

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt[\Large 4]{1+x^2}-\sqrt[\Large 4]{1}}{x^2+1-1}}}$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \displaystyle \large \lim_{x\,\to\,0}{\normalsize \dfrac{\sqrt[\Large 4]{1+x^2}-\sqrt[\Large 4]{1}}{1+x^2-1}}}$

$(1).\,\,\,$ If $x \,\to\,0$, then $x^2 \,\to\,0^2$. Therefore, $x^2 \,\to\,0$

$(2).\,\,\,$ If $x^2 \,\to\,0$, then $1+x^2 \,\to\,1+0$. Therefore, $1+x^2 \,\to\,0$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \displaystyle \large \lim_{1+x^2\,\to\,1}{\normalsize \dfrac{\sqrt[\Large 4]{1+x^2}-\sqrt[\Large 4]{1}}{1+x^2-1}}}$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \displaystyle \large \lim_{1+x^2\,\to\,1}{\normalsize \dfrac{(1+x^2)^{\Large \frac{1}{4}}-(1)^{\Large \frac{1}{4}}}{1+x^2-1}}}$

For our convenience, take $m = x^2$ and then express the whole mathematical expression in terms of $m$.

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \displaystyle \large \lim_{m\,\to\,1}{\normalsize \dfrac{m^{\Large \frac{1}{4}}-(1)^{\Large \frac{1}{4}}}{m-1}}}$

We can now evaluate the limit of the rational expression by the quotient difference limit rule in exponential form.

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \dfrac{1}{4} \times (1)^{\Large \frac{1}{4} \normalsize -1}}$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \dfrac{1}{4} \times (1)^{\Large \frac{1-1\times 4}{4}}}$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \dfrac{1}{4} \times (1)^{\Large \frac{1-4}{4}}}$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \dfrac{1}{4} \times (1)^{\Large \frac{-3}{4}}}$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \dfrac{1}{4} \times 1}$

$=\,\,\,$ $-2 \times \dfrac{1}{4 \times \dfrac{1}{4}}$

$=\,\,\,$ $-2 \times \dfrac{1}{\dfrac{4 \times 1}{4}}$

$=\,\,\,$ $-2 \times \dfrac{1}{\dfrac{4}{4}}$

$=\,\,\,$ $\require{cancel} -2 \times \dfrac{1}{\dfrac{\cancel{4}}{\cancel{4}}}$

$=\,\,\,$ $-2 \times \dfrac{1}{1}$

$=\,\,\,$ $-2 \times 1$