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Suppose $f\left ( x \right )$ is differentiable at $x= 1 \: and\: \lim_{h\rightarrow 0}\frac{1}{h}f\left ( 1+h \right )= 5,$

then ${f}'\left ( 1 \right )$ equals

Option 1)
$4$

Option 2)
$3$

Option 3)
$6$

Option 4)
$5$

Answers (1)

best_answer

As we learnt in

Differentiability -

Let f(x) be a real valued function defined on an open interval (a, b) and $x\epsilon$ (a, b).Then the function f(x) is said to be differentiable at $x_{\circ }$ if

$\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}$

$or\:\:\:\lim_{h\rightarrow 0}\:\frac{f(x)-f(x_{0})}{x-x_{0}}$

-

$f{}'\left ( 1 \right )=\lim_{h\rightarrow 0}\frac{f\left ( 1+h \right )-f\left ( 1 \right )}{h}$

$\lim_{h\rightarrow 0}\frac{f\left ( 1+h \right )}{h}-\lim_{h\rightarrow 0}\frac{f\left ( 1 \right )}{h}$

Now $\lim_{h\rightarrow 0}\frac{f\left ( 1+h \right )}{h}=5$

so that $\lim_{h\rightarrow 0}\frac{f\left ( 1 \right )}{h}$ must be finite as $f{}'\left ( 1 \right )$ exists But $\lim_{h\rightarrow 0}\frac{f\left ( 1 \right )}{h}$ can be finite only if $f\left ( 1 \right )=0$ and $\lim_{h\rightarrow 0}\frac{f\left ( 1 \right )}{h}=0$

Therefore $f{}'\left ( 1 \right )=\lim_{h\rightarrow 0}\frac{f\left ( 1+h \right )}{h}=5$

Option 1)

$4$

Incorrect option

Option 2)

$3$

Incorrect option

Option 3)

$6$

Incorrect option

Option 4)

$5$

Correct option

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prateek

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