Evaluate the limit: lim _{x rightarrow 1} lef | vedclass

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Question

(A) Given the limit: $\lim _{x \rightarrow 1} \left[\frac{\sqrt{x}-1}{\log x}\right]$ Since the form is $\frac{0}{0}$ as $x \rightarrow 1$,we apply $L

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) Given the limit: $\lim _{x ightarrow 1} \left[\frac{\sqrt{x}-1}{\log x}ight]$ Since the form is $\frac{0}{0}$ as $x ightarrow 1$,we apply $L'	ext{Hospital's rule}$ by differentiating the numerator and denominator with respect to $x$: $= \lim _{x ightarrow 1} \frac{\frac{d}{dx}(\sqrt{x}-1)}{\frac{d}{dx}(\log x)}$ $= \lim _{x ightarrow 1} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{x}}$ $= \lim _{x ightarrow 1} \frac{x}{2\sqrt{x}} = \lim _{x ightarrow 1} \frac{\sqrt{x}}{2}$ $= \frac{\sqrt{1}}{2} = \frac{1}{2}$

Evaluate the limit: $\lim _{x \rightarrow 1} \left[\frac{\sqrt{x}-1}{\log x}\right]$

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