Let f : (0, ) (2,20) be twice differentiable function such that _… - Vidyadip

MCQ

Let $f : (0, \infty) \to (2,20)$ be twice differentiable function such that $\mathop {\lim }\limits_{x \to \infty } (f(x) + f'(x) + f^{''}(x)) = \mathop {\lim }\limits_{x \to \infty } g(x)$

where $\mathop {\lim }\limits_{x \to \infty } g(x)$ exists and equal to $5$, then $\mathop {\lim }\limits_{x \to \infty } (f(x) - g(x))$ equal to

A
$5$
B
$7$
✓
$0$
D
Does not exists

✓

Answer

Correct option: C.

$0$

c
Given $\mathop {\lim }\limits_{x \to \infty } \left( {f(x) + {f^\prime }(x) + {f^{\prime \prime }}(x)} \right) = 5$

$\therefore \mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } \frac{{f(x) \cdot {e^x}}}{{{e^x}}}$

$ = \mathop {\lim }\limits_{x \to \infty } \frac{{\left( {f(x) + {f^\prime }(x)} \right){e^x}}}{{{e^x}}}$

(By L'Hópital rule)

$ = \mathop {\lim }\limits_{x \to \infty } \left( {f(x) + \frac{{{f^\prime }(x){e^x}}}{{{e^x}}}} \right)$

$ = \mathop {\lim }\limits_{x \to \infty } \left( {f(x) + \frac{{\left( {{f^\prime }(x) + {f^{\prime \prime }}(x)} \right){e^x}}}{{{e^x}}}} \right)$

$ = \mathop {\lim }\limits_{x \to \infty } \left( {f(x) + {f^\prime }(x) + {f^\prime }(x)} \right) = 5$

$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } (f(x) - g(x)) = 0$

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