MCQ
$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{1/x}} - 1}}{{{e^{1/x}} + 1}} = $
- A$0$
- B$1$
- C$-1$
- ✓અસ્તિત્વ નથી.
$\mathop {\lim }\limits_{x \to \,0 + } \,f(x) = \mathop {\lim }\limits_{h \to \,0} \left( {\frac{{{e^{1/h}} - 1}}{{{e^{1/h}} + 1}}} \right) $
$= \mathop {\lim }\limits_{h \to \,0} \frac{{{e^{1/h}}\left( {1 - \frac{1}{{{e^{1/h}}}}} \right)}}{{{e^{1/h}}\left( {1 + \frac{1}{{{e^{1/h}}}}} \right)}} = 1$
Similarly $\mathop {\lim }\limits_{x \to \,0 - } f(x) = - 1$.
Hence limit does not exist.
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