If f is strictly increasing function, then _ x 0 f( x^2 ) - f(x) … - Vidyadip

MCQ

If $f$ is strictly increasing function, then $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$ is equal to

A
$0$
B
$1$
✓
$-1$
D
$2$

✓

Answer

Correct option: C.

$-1$

c
(c) $\mathop {\lim }\limits_{x \to 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$, $\left( {\frac{0}{0} \,\, {\rm{form}}} \right)$

$ = \mathop {\lim }\limits_{x \to 0} \frac{{2xf'({x^2}) - f'(x)}}{{f'(x)}}$, (using $L'$ Hospital's rule)

$ = - 1 + \mathop {\lim }\limits_{x \to 0} \frac{{2xf'({x^2})}}{{f'(x)}} = - 1,f'(0) \ne 0,\,$

as $f$ is strictly increasing.

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