Search any question & find its solution
Question:
Answered & Verified by Expert
If $\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x}$ exists and is equal to 1 then the value of $a$ is
Options:
Solution:
2443 Upvotes
Verified Answer
The correct answer is:
1
$\lim _{x \rightarrow 0} \frac{2 a \sin x-\sin 2 x}{\tan ^{3} x}$
$$
=\lim _{x \rightarrow 0} \frac{2 a\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\ldots\right)-\left(2 x-\frac{(2 x)^{3}}{3 !}+\frac{(2 x)^{5}}{5 !}-\ldots\right)}{\left(x+\frac{x^{3}}{3}+\frac{2}{15} x^{5}+\ldots\right)^{3}}
$$
$$
=\lim _{x \rightarrow 0} \frac{\left(2 a-2 x+\left(-\frac{2 a}{3 !}+\frac{2^{3}}{3 !}\right) x^{3}+\left(\frac{2 a}{5 !}-\frac{2^{5}}{5 !}\right) x^{5}+\ldots\right.}{x^{3}\left(1+\frac{x^{2}}{3}+\frac{2}{15} x^{4}+\ldots\right)^{3}}
$$
since, it is given that given limit is exist, then
$$
2 a-2=0 \Rightarrow \alpha=1
$$
$$
=\lim _{x \rightarrow 0} \frac{2 a\left(x-\frac{x^{3}}{3 !}+\frac{x^{5}}{5 !}-\ldots\right)-\left(2 x-\frac{(2 x)^{3}}{3 !}+\frac{(2 x)^{5}}{5 !}-\ldots\right)}{\left(x+\frac{x^{3}}{3}+\frac{2}{15} x^{5}+\ldots\right)^{3}}
$$
$$
=\lim _{x \rightarrow 0} \frac{\left(2 a-2 x+\left(-\frac{2 a}{3 !}+\frac{2^{3}}{3 !}\right) x^{3}+\left(\frac{2 a}{5 !}-\frac{2^{5}}{5 !}\right) x^{5}+\ldots\right.}{x^{3}\left(1+\frac{x^{2}}{3}+\frac{2}{15} x^{4}+\ldots\right)^{3}}
$$
since, it is given that given limit is exist, then
$$
2 a-2=0 \Rightarrow \alpha=1
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.