MCQ
x→0lim \Bigg(\dfrac{(1+x)^{2}}{e^{x}}\Bigg)^\dfrac{4}{\sin x}(ex(1+x)2)sinx4 is:
- A$ \text{e}^2$
- B$ \text{e}^4$
- C$ \text{e}^8$
- D$ \text{e}^9$
Solution:
$ \lim\limits_{\text{x} \to 0}\left(\frac{(1+\text{x})^2}{\text{e}^{\text{x}}}\right)^{\frac{4}{\sin \text{x}}}=\lim\limits_{\text{x} \to 0}\frac{\left(\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\frac{\text{x}}{\sin \text{x}}}\right)^8}{\text{e}^{\frac{4\text{x}}{\sin \text{x}}}}$
We have
$ \lim\limits_{\text{x} \to 0}\frac{\sin \text{x}}{\text{x}}=1$
and $ \lim\limits_{\text{x} \to 0}(1+\text{x})^{\frac{1}{\text{x}}}=$
both the limits of the numerator and denominator exists,and the limit of the numerator is non-vanishing,
$=\lim_\limits{\text{x} \rightarrow 0}\frac{\left(\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\lim_\limits{\text{x} \rightarrow 0}\frac{\text{x}}{\sin \text{x}}}\right)^8}{\lim_\limits{\text{x} \rightarrow 0}\text{e}^{\frac{4\text{x}}{\sin \text{x}}}}$
[Using division property of limits]
$=\frac{\left(\lim\limits_{\text{x}\to 0}\left\{(1+\text{x})^{\frac{1}{\text{x}}}\right\}^{\left(\lim\limits_{\text{x}\to 0}\frac{\text{x}}{\sin \text{x}}\right)}\right)^8}{\text{e}^{\left(\lim\limits_{\text{x}\to 0}\dfrac{4\text{x}}{\sin \text{x}}\right)}}$ [Using limit property]
$$$= \text{e}^4$
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