Question

Solve:
$\underset{x\to 0}{lim}$ $(\frac{1}{x}-\frac{\log (1+x)}{x^2})$

Answer 1

We have,

$\underset{x\to 0}{lim}(\frac{1}{x}-\frac{\log (1+x)}{x^2})$

$\underset{x\to 0}{lim}\left(\frac{x-\log (1+x)}{x^2}\right)$

We know that

$\log (1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-......$

Therefore,

$\underset{x\to 0}{lim}\left(\frac{x-(x-\frac{x^2}{2}+\frac{x^3}{3}-.....)}{x^2}\right)$

$\underset{x\to 0}{lim}\left(\frac{x-x+\frac{x^2}{2}-\frac{x^3}{3}+.....)}{x^2}\right)$

$\underset{x\to 0}{lim}\left(\frac{\frac{x^2}{2}-\frac{x^3}{3}+.....)}{x^2}\right)$

$\underset{x\to 0}{lim}(x^2\times \frac{(\frac{1}{2}-\frac{x}{3}+.....)}{x^2})$

$\underset{x\to 0}{lim}(\frac{1}{2}-\frac{x}{3}+.....)$

$=\frac{1}{2}-\frac{0}{3}+0-.....$

$=\frac{1}{2}$

Hence, this is the answer.