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Question

(D) Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{1/x}}}}{{{e^{\left( {\frac{1}{x} + 1} ight)}}}}$Using the property of exponents,$e

Vedclass · Accepted Answer

None of these

Vedclass · Answer

(D) Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{1/x}}}}{{{e^{\left( {\frac{1}{x} + 1} ight)}}}}$Using the property of exponents,$e^{a+b} = e^a \cdot e^b$,we can rewrite the denominator:$\mathop {\lim }\limits_{x 	o 0} \frac{{{e^{1/x}}}}{{{e^{1/x}} \cdot e^1}}$Canceling the common term $e^{1/x}$ (which is non-zero for all $x 
eq 0$):$\mathop {\lim }\limits_{x 	o 0} \frac{1}{e} = \frac{1}{e} = e^{-1}$Since $e^{-1}$ is not equal to $0$,$1$,or does not exist,the correct option is $D$.

$\mathop {\lim }\limits_{x \to 0} \frac{{{e^{\frac{1}{x}}}}}{{{e^{\left( {\frac{1}{x} + 1} \right)}}}} = $

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