Limit Continuity And Differentiability Ques 79
- $\lim _{x \rightarrow 0} \frac{(1-\cos 2 x)(3+\cos x)}{x \tan 4 x}$ is equal to
(2013 Main)
4
3
2
(d) $\frac{1}{2}$
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Answer:
Correct Answer: 79.(c)
Solution:
- Here, $\lim _{x \rightarrow 0}{1+\underset{\substack{a \rightarrow 0}}{ }\left(\log \left(1+b^{2}\right) }^{1 / x}\right.$
$\left[1^{\infty}\right]$ form
$$ \begin{alignedat} & =e^{\lim _{x \rightarrow 0}\left(x \log \left(1+b^{2}\right) \right) \cdot \frac{1}{x}} \\ & =e^{\log \left(1+b^{2}\right)}=\left(1+b^{2}\right) \end{aligned} $$
Given, $\lim _{x \rightarrow 0}{1+x \log \left(1+b^{2}\right) }^{1 / x}=e^{ \log \left(1+b^{2}\right)}$
$$ \Rightarrow \quad x \rightarrow 0 \quad\left(1+b^{2}\right)=2 b \sin ^{2} \theta $$
$$ \therefore \quad \sin ^{2} \theta=\frac{1+b^{2}}{2 b} $$
By AM $\geq$ GM, $\frac{b+\frac{1}{b}}{2} \geq \sqrt{b \cdot \frac{1}{b}}$
$$ \Rightarrow \quad \frac{b^{2}+1}{2 b} \geq 1 $$
From Eqs. (2) and (3),
$$ \begin{aligned} & \sin ^{2} \theta + \cos ^{2} \theta = 1 \\ & \Rightarrow \quad \theta= \pm \frac{\pi}{2} \text {, as } \theta \in(-\pi, \pi] \end{aligned} $$
BETA
