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