JEE MAIN - Mathematics (2005 - No. 4)
Suppose $$f(x)$$ is differentiable at x = 1 and
$$\mathop {\lim }\limits_{h \to 0} {1 \over h}f\left( {1 + h} \right) = 5$$, then $$f'\left( 1 \right)$$ equals
$$\mathop {\lim }\limits_{h \to 0} {1 \over h}f\left( {1 + h} \right) = 5$$, then $$f'\left( 1 \right)$$ equals
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Обяснение
$$f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h};$$
As function is differentiable so it is continuous as it
is given that $$\mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5$$ and hence $$f(1)=0$$
Hence $$f'(1)$$ $$ = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5$$
As function is differentiable so it is continuous as it
is given that $$\mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5$$ and hence $$f(1)=0$$
Hence $$f'(1)$$ $$ = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5$$