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If $f(x)=\frac{\sin \left(e^{x-2}-1\right)}{\ln (x-1)}$, then $\lim _{x \rightarrow 2} f(x)$ is equal to
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$\mathrm{f}(\mathrm{x})=\frac{\sin \left(\mathrm{e}^{\mathrm{x}-2}-1\right)}{\ln (\mathrm{x}-1)}$
$\lim _{\mathrm{x} \rightarrow 2} \frac{\sin \left(\mathrm{e}^{\mathrm{x}-2}-1\right)}{\ln (\mathrm{x}-1)}=\mathrm{L}$
It is $\frac{0}{0}$ (undefined) condition so using L'hospital's rule
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2}\left[\frac{\left\{\sin \left(\mathrm{e}^{\mathrm{x}-2}-1\right)\right\}^{-}}{\{\ln (\mathrm{x}-1)\}^{-}}\right]$
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2} \frac{\cos \left(\mathrm{e}^{\mathrm{x}-2}-1\right) \cdot \mathrm{e}^{(\mathrm{x}-2)}}{1 /(\mathrm{x}-1)}$
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2} \cos \left(\mathrm{e}^{2-2}-1\right) \mathrm{e}^{2-2} \cdot(2-1)$
$\Rightarrow \mathrm{L}=\cos (0) \mathrm{e}^{0} \cdot 1$
$\Rightarrow \mathrm{L}=1$
$\lim _{\mathrm{x} \rightarrow 2} \frac{\sin \left(\mathrm{e}^{\mathrm{x}-2}-1\right)}{\ln (\mathrm{x}-1)}=\mathrm{L}$
It is $\frac{0}{0}$ (undefined) condition so using L'hospital's rule
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2}\left[\frac{\left\{\sin \left(\mathrm{e}^{\mathrm{x}-2}-1\right)\right\}^{-}}{\{\ln (\mathrm{x}-1)\}^{-}}\right]$
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2} \frac{\cos \left(\mathrm{e}^{\mathrm{x}-2}-1\right) \cdot \mathrm{e}^{(\mathrm{x}-2)}}{1 /(\mathrm{x}-1)}$
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2} \cos \left(\mathrm{e}^{2-2}-1\right) \mathrm{e}^{2-2} \cdot(2-1)$
$\Rightarrow \mathrm{L}=\cos (0) \mathrm{e}^{0} \cdot 1$
$\Rightarrow \mathrm{L}=1$
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