If $\lim _{x \rightarrow 0^{+}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=k$ and…

If $\lim _{x \rightarrow 0^{+}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=k$ and $\lim _{x \rightarrow 0^{-}} x^2\left(\frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\right)=l$, then
  1. $k=1$
  2. $k=1, I=-1$
  3. $k=-1, /=1$
  4. $k \neq I \neq \pm 1$

Solution

$k=\lim _{x \rightarrow 0^{+}} x^2\left[\frac{e^{\frac{1}{x}}-e^{\frac{-1}{x}}}{e^{\frac{1}{x}}+e^{\frac{-1}{x}}}\right]$ Let $\frac{1}{x}=t$ $x \rightarrow 0^{+}, t \rightarrow \infty$ $k=\lim _{t \rightarrow \infty} \frac{1}{t^2}\left[\frac{e^t-e^{-t}}{e^t+e^{-t}}\right]$ $=\lim _{x \rightarrow \infty} \frac{1}{t^2}\left[\frac{1-e^{-2 t}}{1+e^{-2 t}}\right]$=0$ $\left[\begin{array}{l}\because \lim _{t \rightarrow \infty} \frac{1}{e^{2 t}}=0 \\ \text { and } \lim _{t \rightarrow \infty} \frac{1}{t^2}=0\end{array}\right]$ $l=\lim _{x \rightarrow 0^{-}} x^2\left[\frac{e^{\frac{1}{x}}-e^{-\frac{1}{x}}}{e^{\frac{1}{x}}+e^{-\frac{1}{x}}}\right]$ Let $\frac{1}{x}=y$ $x \rightarrow 0^{-}, y \rightarrow-\infty$ $\therefore \quad l=\frac{\lim }{y \rightarrow \infty} \frac{1}{y^2}\left[\frac{e^y-e^{-y}}{e^y+e^{-y}}\right]=0 \quad$ [same as above] $\therefore \quad k=l$

Asked in: AP EAMCET 2022 (07 Jul Shift 1)

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