Evaluate the following limits: _ x 1 [ x-1 x ]

Question

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

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Answer

$\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{\log x}$
put $x-1=\mathrm{h}$
$\therefore \quad x=1+\mathrm{h}$
As $x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$\therefore$ Required limit
$=\lim _{h \rightarrow 0} \frac{\sqrt{1+h}-1}{\log (1+h)}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{\sqrt{1+\mathrm{h}}-1}{\log (1+\mathrm{h})} \times \frac{\sqrt{1+\mathrm{h}}+1}{\sqrt{1+\mathrm{h}}+1} \ldots [$ By rationalization $]$
$=\lim _{h \rightarrow 0} \frac{(1+h)-1}{\log (1+h)} \times \frac{1}{\sqrt{1+h}+1}$
$=\lim _{h \rightarrow 0} \frac{h}{\log (1+h)} \times \frac{1}{\sqrt{1+h}+1}$
$=\lim _{h \rightarrow 0} \frac{1}{\frac{\log (1+h)}{h}} \times \frac{1}{\sqrt{1+h}+1}$
$\ldots\left[\begin{array}{l} \text { Divide Numerator and } \\ \text { Denominator by } \\
\text { Ash } \rightarrow 0, h \neq 0
\end{array} \right]$
$=\lim _{h \rightarrow 0} \frac{1}{\frac{\log (1+h)}{h}} \times \frac{1}{\lim _{h \rightarrow 0}(\sqrt{1+h}+1)}$
$=\frac{1}{1} \times \frac{1}{\sqrt{1+0}+1}$
$=\frac{1}{2}$

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