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If $\mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{5 / \mathrm{x}}$ is continuous at $\mathrm{x}=0$, then what is the value of $\mathrm{f}(0)$ ?
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The correct answer is:
$\mathrm{e}^{5}$
Given that $\mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{5 / \mathrm{x}}$ and $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$. Value of function at $\mathrm{x}=0$ is same as limit of the function at $\mathrm{x}=0$.
$\mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0}(1+\mathrm{x})^{5 / \mathrm{x}}=\left\{\lim _{\mathrm{x} \rightarrow 0}(1+\mathrm{x})^{1 / \mathrm{x}}\right\}^{5}=\mathrm{e}^{5}$
$\mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0}(1+\mathrm{x})^{5 / \mathrm{x}}=\left\{\lim _{\mathrm{x} \rightarrow 0}(1+\mathrm{x})^{1 / \mathrm{x}}\right\}^{5}=\mathrm{e}^{5}$
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