GATE CSE 2022 | Question: 24

Question

The value of the following limit is ________________.

$$\lim_{x \rightarrow 0^{+}} \frac{\sqrt{x}}{1-e^{2\sqrt{x}}}$$

Comments

A Property of Limits and a standard limit:-

$1)$ $ $ if $\lim_{x \to c } f(x) = L$ then $\lim_{x \to c } \dfrac{1}{f(x)} = \dfrac{1}{L}$ $ $ (Provided that $L \neq 0$).

$2)$ $\lim_{x \to 0} \dfrac{a^x  -1 }{x} = \ln(a)$. [Applicable for Right and left hand limits also].

Given limit:- $\lim_{x \to 0^+} \dfrac{\sqrt{x}}{1-e^{2\sqrt{x}}} = -\lim_{x \to 0^+} \dfrac{\sqrt{x}}{e^{2\sqrt{x}} -1}$

Now Divide numerator and denominator with $2\sqrt{x}$, then it will be,

$= -\lim_{x \to 0^+} \dfrac{\frac{1}{2}}{\frac{e^{2\sqrt{x}} -1}{2\sqrt{x}}}$

$= -\dfrac{1}{2}\lim_{x \to 0^+} \dfrac{1}{\frac{e^{2\sqrt{x}} -1}{2\sqrt{x}}}$

Now Evaluate the limit in denominator part, $\lim_{x \to 0^+}\dfrac{e^{2\sqrt{x}}-1}{2\sqrt{x}}$

Let $y = 2\sqrt{x}$, as $x \to 0^+, y \to 0^+$, we i rewrite it as ,

$=\lim_{y \to 0^+}\dfrac{e^y-1}{y}$ (which is in form of standard which mentioned above)

$=\ln(e) = 1$

so $ -\dfrac{1}{2}\lim_{x \to 0^+} \dfrac{1}{\frac{e^{2\sqrt{x}} -1}{2\sqrt{x}}} = -\dfrac{1}{2} \dfrac{\lim_{x \to 0^+} 1 }{\lim_{x \to 0^+}\dfrac{e^{2\sqrt{x}}-1}{2\sqrt{x}}} = -\dfrac{1}{2} (1) = -\dfrac{1}{2}$.
 

Method 2 :- Apply L'Hopital Rule, Method 3:-  Series Expansion.

Answer 1

Given,

$\lim_{x \rightarrow 0^{+}} \frac{\sqrt{x}}{1-e^{2\sqrt{x}}}$ $(0/0  \ form)$

Using L'hopital rule

$\displaystyle{}\lim_{x\to0} \frac{\frac{1}{2\sqrt(x)}}{-e^{2\sqrt(x)} \times \frac{2}{2\sqrt(x)}} = \lim_{x\to0} \frac{1}{-2e^{2\sqrt(x)}} = -\frac{1}{2} = -0.5$

Comments

Thanks.

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@Sachin Mittal 1 sir, if we apply (1-a^x)/x format we get 1/2(ln e) = 0.5 which is wrong, why is this so?

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Standard limit is:- $\large{\lim_{x \to 0} \dfrac{a^x -1}{x} =0}$,  for the given question Frist you have to factor out $\large{-1}$ from the denominator part, to convert into standard form.

Answer 2

Answer is -0.5 .

Answer 3

$\displaystyle \lim_{x \to 0+} \frac{\sqrt{x}}{1-e^{2\sqrt{x}}}\; (\frac{0}{0}\,form)$

$\text{Using L'Hôpital's rule}$

$\displaystyle \lim_{x \to 0+} \frac{\frac{1}{2\sqrt{x}}}{0-e^{2\sqrt{x}}.\frac{2}{2\sqrt{x}}} =\lim_{x \to 0+} -\frac{1}{2.e^{2\sqrt{x}}} =-\frac{1}{2} =-0.5$

Answer 4

Using, $e^x = \Sigma_{n=0}^{\infty}\frac{x^n}{n!}$ (Taylor Series Expansion for $e^x$ near $x=0$)

$\lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{1-\left(1 + \frac{2\sqrt{x}}{1!} + \frac{(2\sqrt{x})^2}{2!} +... \right )} = \lim_{x \rightarrow 0^+} \frac{\sqrt{x}}{-\left( \frac{2\sqrt{x}}{1!} + \frac{(2\sqrt{x})^2}{2!} +... \right )}$

$= \lim_{x \rightarrow 0^+} \frac{-1}{\left( \frac{2}{1!} + \frac{2^2\sqrt{x}}{2!} +... \right )}, x \neq 0$

$ = \frac{-1}{2}$