Evaluate $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}}$

The function is an algebraic function and its limit can be calculated by using direct substitution method.

$\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2}}$

$= \,\,\,$ $\dfrac{\sqrt{1+2(4)}-3}{\sqrt{4}-2}$

$= \,\,\,$ $\dfrac{\sqrt{1+8}-3}{2-2}$

$= \,\,\,$ $\dfrac{\sqrt{9}-3}{2-2}$

$= \,\,\,$ $\dfrac{3-3}{2-2}$

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

It is an indeterminate form and it’s not possible to evaluate the limit of this algebraic function by the direct substitution method.

Try Rationalization method

The numerator and denominator both are in radical form. They should be free from this form and it can be done by rationalizing both numerator and denominator by their conjugate functions. Let’s first apply rationalisation to algebraic function in the numerator and then apply to denominator.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2} \times 1 \Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{\sqrt{1+2x}-3}{\sqrt{x}-2} \times \dfrac{\sqrt{1+2x}+3}{\sqrt{1+2x}+3} \Bigg]}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{(\sqrt{1+2x}-3)(\sqrt{1+2x}+3)}{\sqrt{x}-2}}$ $\times$ $\dfrac{1}{\sqrt{1+2x}+3} \normalsize \Bigg]$

According to difference of squares property, the special product of binomials can be simplified as difference of squares of their terms.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{{(\sqrt{1+2x})}^2-3^2}{\sqrt{x}-2}}$ $\times$ $\dfrac{1}{\sqrt{1+2x}+3} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{1+2x-9}{\sqrt{x}-2}}$ $\times$ $\dfrac{1}{\sqrt{1+2x}+3} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2x+1-9}{\sqrt{x}-2}}$ $\times$ $\dfrac{1}{\sqrt{1+2x}+3} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2x-8}{\sqrt{x}-2}}$ $\times$ $\dfrac{1}{\sqrt{1+2x}+3} \normalsize \Bigg]$

The algebraic function in the denominator of the first factor made the entire function undefined as $x$ tends to $4$. So, use rationalizing technique to rationalise the denominator.

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2x-8}{\sqrt{x}-2} \times 1}$ $\times$ $\dfrac{1}{\sqrt{1+2x}+3} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2x-8}{\sqrt{x}-2}}$ $\times$ $\dfrac{\sqrt{x}+2}{\sqrt{x}+2}$ $\times$ $\dfrac{1}{\sqrt{1+2x}+3} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2x-8}{(\sqrt{x}-2)(\sqrt{x}+2)}}$ $\times$ $\dfrac{(\sqrt{x}+2) \times 1}{\sqrt{1+2x}+3} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2x-8}{{(\sqrt{x})}^2-2^2}}$ $\times$ $\dfrac{\sqrt{x}+2}{\sqrt{1+2x}+3} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2x-8}{x-4}}$ $\times$ $\dfrac{\sqrt{x}+2}{\sqrt{1+2x}+3} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2(x-4)}{x-4} \times \dfrac{\sqrt{x}+2}{\sqrt{1+2x}+3}} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \require{cancel} \lim_{x \,\to\, 4}{\normalsize \Bigg[\dfrac{2\cancel{(x-4)}}{\cancel{x-4}} \times \dfrac{\sqrt{x}+2}{\sqrt{1+2x}+3}} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \Bigg[2 \times \dfrac{\sqrt{x}+2}{\sqrt{1+2x}+3}} \normalsize \Bigg]$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \dfrac{2 \times (\sqrt{x}+2)}{\sqrt{1+2x}+3}}$

$= \,\,\,$ $\displaystyle \large \lim_{x \,\to\, 4}{\normalsize \dfrac{2(\sqrt{x}+2)}{\sqrt{1+2x}+3}}$