If f is a strictly increasing function,then m | vedclass

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(C) Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$.This is a $\frac{0}{0}$ indeterminate form.Applying $L

Vedclass · Accepted Answer

$-1$

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(C) Given the limit: $\mathop {\lim }\limits_{x 	o 0} \frac{{f({x^2}) - f(x)}}{{f(x) - f(0)}}$.This is a $\frac{0}{0}$ indeterminate form.Applying $L'	ext{Hospital's rule}$,we differentiate the numerator and denominator with respect to $x$:$= \mathop {\lim }\limits_{x 	o 0} \frac{{2x \cdot f'({x^2}) - f'(x)}}{{f'(x)}}$.$= \mathop {\lim }\limits_{x 	o 0} \left( \frac{2x \cdot f'({x^2})}{f'(x)} - \frac{f'(x)}{f'(x)} ight)$.$= \mathop {\lim }\limits_{x 	o 0} \left( \frac{2x \cdot f'({x^2})}{f'(x)} - 1 ight)$.Since $f$ is strictly increasing,$f'(x) > 0$. Assuming $f'(0) 
eq 0$,the term $\mathop {\lim }\limits_{x 	o 0} \frac{2x \cdot f'({x^2})}{f'(x)} = \frac{0 \cdot f'(0)}{f'(0)} = 0$.Therefore,the limit is $0 - 1 = -1$.

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

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