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Limits — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
Question
Show that $\lim\limits_{\text{x}\rightarrow2}\text{ e}^\frac{-1}{\text{x}}$ does not exist.
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Answer

$\lim\limits_{\text{x}\rightarrow0}\text{ e}^{\frac{-1}{​​\text{x}}}$
$\lim\limits_{\text{x}\rightarrow0^+}\text{ e}^{\frac{-1}{​​\text{x}}}=\lim\limits_{\text{x}\rightarrow0^+}\frac{1}{\text{e}^{\frac{1}{\text{x}}}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{e}^{\frac{1}{0+\text{h}}}}=\frac{1}{\text{e}^{\frac{1}{0}}}=\frac{1}{\text{e}^{\infty}}=\frac{1}{\infty}=0$
And,
$=\lim\limits_{\text{x}\rightarrow0^-}\text{e}^{\frac{-1}{\text{x}}}=\lim\limits_{\text{h}\rightarrow0^-}\frac{1}{\text{e}^{\frac{1}{\text{x}}}}$
$=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{e}^\frac{1}{0-\text{h}}}=\lim\limits_{\text{h}\rightarrow0}\frac{1}{\text{e}^-\frac{1}{\text{h}}}=\frac{1}{\text{e}^-\frac{1}{0}}=\frac{1}{\text{e}^{-\infty}}=\text{e}^\infty=\infty$
$\Rightarrow\lim\limits_{\text{x}\rightarrow0^+}\text{e}^{\frac{-1}{\text{x}}}\ne\lim\limits_{\text{x}\rightarrow0^-}\text{e}^{\frac{-1}{\text{x}}}$
$\therefore\ \lim\limits_{\text{x}\rightarrow0}\text{e}^{\frac{-1}{\text{x}}}$ does not exist.

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