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L'Hospital Rule to Remove Indeterminate Form

Let f:R→ R ...

Question

Let $f : R \to R$ be a differentiable function and $f (1) = 4.$ Then the value of $lim x \to 1 \int_{4}^{f (x)} \frac{2 t}{x - 1} d t$ is

A

8 f^{'} (1)

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B

4 f^{'} (1)

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C

2 f^{'} (1)

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D

f^{'} (1)

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Solution

The correct option is A $8 f^{'} (1)$
${lim}_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t = {lim}_{x \to 1} \frac{\int_{4}^{f (x)} 2 t d t}{x - 1}$
Applying L'Hospital's rule
$= {lim}_{x \to 1} \frac{2 f (x) f^{'} (x)}{1} = f (1) f^{'} (1) = 8 f^{'} (1)$

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Similar questions

\to

R is differentiable function &

f (1) = 4

g (x) = lim x \to 1 \int_{4}^{f (x)} \frac{2 t}{x - 1} d t

f (x)

be a differentiable function and

f (1) = 2

lim x \to 1 \int_{2}^{f (x)} \frac{2 t}{x - 1} d t = 4

, then the value of

f^{'} (1)

f :

R \to R

be a differentiable function at

x = 1

f (1) = 4

f^{'} (1) = 2

α = {lim}_{x \to 1} \int_{4}^{f (x)} \frac{2 t}{x - 1} d t

α

f : R \to R

be such that f(1) =2, f'(1)= 4 then the

{lim}_{x \to 0} [\frac{f (1 + x + x^{2}) + x f (1 + x)}{f (1)}]^{\frac{1}{x (x + 1)}}

F : R \to R

be a differentiable function such that

F (x) = \int_{4}^{f (x)} \frac{16 t^{3}}{x - 3} d t, f (3) = 4 & f^{'} (3) = \frac{1}{16}

then the value of

lim x \to 3 F (x)

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