Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2^x-1}{\sqrt{1+x}-1}}$

Evaluate the limit of the exponential algebraic function as $x$ approaches zero by the direct substitution method.

$= \,\,\,$ $\dfrac{2^{0}-1}{\sqrt{1+0}-1}$

$= \,\,\,$ $\dfrac{1-1}{\sqrt{1}-1}$

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

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

In this limit problem, the limit of the exponential algebraic function is undefined as $x$ approaches zero. So, the direct substitution method is not useful in this case and we have to think about this limit problem in another way.

Find the Limit of Exponential function

$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{2^x-1}{\sqrt{1+x}-1}}$

The expression in the numerator is similar to limit of $\dfrac{a^x-1}{x}$ as $x$ approaches $0$ formula but its denominator should have $x$. So, try to make the numerator to have $x$ as its denominator.

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

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

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

Use product rule of limits to find the limit of the factors by finding the product of their limits.

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

According to limit of (ax-1)/x as x approaches 0 exponential rule, the limit of the exponential function is equal to natural logarithm of $2$.

$= \,\,\,$ $\log_{e}{(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x}{\sqrt{1+x}-1}}$

$= \,\,\,$ $\ln{(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x}{\sqrt{1+x}-1}}$

Find the Limit of Algebraic function

Now, find the limit of algebraic function by the direct substitution method.

$= \,\,\,$ $\ln{(2)}$ $\times$ $\dfrac{0}{\sqrt{1+0}-1}$

$= \,\,\,$ $\ln{(2)}$ $\times$ $\dfrac{0}{\sqrt{1}-1}$

$= \,\,\,$ $\ln{(2)}$ $\times$ $\dfrac{0}{1-1}$

$= \,\,\,$ $\ln{(2)}$ $\times$ $\dfrac{0}{0}$

The limit of the function is undefined as $x$ approaches zero. So, the direct substitution method should not be applied for this function and try alternative method.

Evaluate the function by Rationalization

$\ln{(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x}{\sqrt{1+x}-1}}$

The algebraic function is in radical form. So, it can be simplified by the rationalisation method.

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

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

$= \,\,\,$ $\ln{(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x{(\sqrt{1+x}+1)}}{{(\sqrt{1+x}-1)}{(\sqrt{1+x}+1)}}}$

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

$= \,\,\,$ $\ln{(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x{(\sqrt{1+x}+1)}}{1+x-1}}$

$= \,\,\,$ $\ln{(2)}$ $\times$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x{(\sqrt{1+x}+1)}}{\cancel{1}+x-\cancel{1}}}$

$= \,\,\,$ $\ln{(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x{(\sqrt{1+x}+1)}}{x}}$

$= \,\,\,$ $\ln{(2)}$ $\times$ $\require{cancel} \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{\cancel{x}{(\sqrt{1+x}+1)}}{\cancel{x}}}$

$= \,\,\,$ $\ln{(2)}$ $\times$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize (\sqrt{1+x}+1)}$

Now, find the limit of the algebraic function in radical form as $x$ approaches zero.

$= \,\,\,$ $\ln{(2)}$ $\times$ $(\sqrt{1+0}+1)$

$= \,\,\,$ $\ln{(2)}$ $\times$ $(\sqrt{1}+1)$

$= \,\,\,$ $\ln{(2)}$ $\times$ $(1+1)$

$= \,\,\,$ $\ln{(2)}$ $\times$ $2$

$= \,\,\,$ $2 \times \ln{(2)}$

$= \,\,\,$ $2\ln{(2)}$

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Jun 26, 2023

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