Limit Continuity and Differentiability 1 Question 10
11. If $\lim _{x \rightarrow \infty} \frac{x^{2}+x+1}{x+1}-a x-b=4$, then
(a) $a=1, b=4$
(b) $a=1, b=-4$
(c) $a=2, b=-3$
(d) $a=2, b=3$
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Answer:
Correct Answer: 11. (b)
Solution:
- Here, $\lim _{x \rightarrow 0}(\sin x)^{1 / x}+\lim _{x \rightarrow 0} \frac{1}{x}^{\sin x}$
$$ =0+\lim _{x \rightarrow 0} e^{\log \frac{1}{x}}=e^{\lim _{x \rightarrow 0} \frac{\log (1 / x)}{\operatorname{cosec} x}} \quad \begin{aligned} & \lim _{x \rightarrow 0}(\sin x)^{1 / x} \rightarrow 0 \\ & \\ & \text { as, }(\text { decimal })^{\infty} \rightarrow 0 \end{aligned} $$
Applying L’Hospital’s rule, we get
$$ e^{\lim _{x \rightarrow 0} \frac{x-\frac{1}{x^{2}}}{-\operatorname{cosec} x \cot x}}=e^{\lim _{x \rightarrow 0} \frac{\sin x}{x} \tan x}=e^{0}=1 $$
