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$2$
$ = \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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