Question

$\underset{x\to 1}{lim}\frac{\log x}{x-1}$ is equal to

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Answer 1

We have, $\underset{x\to 1}{lim}\frac{\log x}{x-1}$

At $x=0$, the value of the expression is the form $\frac{0}{0}$

Applying L'Hospital Rule, we get
$\underset{x\to 1}{lim}\frac{\frac{d}{dx}\log x}{\frac{d}{dx}(x-1)}$

$\Rightarrow \underset{x\to 1}{lim}\frac{\frac{1}{x}}{1}=\underset{x\to 1}{lim}\frac{1}{x}$

Since, $(\frac{d}{dx}\log x=\frac{1}{x})$

Putting $x=1$, we get the value of the limit as $\frac{1}{1}=1$