Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

Evaluate $\lim \limits_{x \to 0} \bigg( \large\frac{\sin 2x+ \sin 6x}{\sin 5x -\sin 3x} \bigg)$

$\begin{array}{1 1}(A)\;4 \\(B)\;2 \\(C)\;3 \\(D)\;0 \end{array}$

Can you answer this question?

1 Answer

0 votes
$\lim \limits_{x \to 0} \bigg( \large\frac{ \sin 2x + \sin 6x}{\sin 5x - \sin 3x} \bigg)$
$\qquad =\lim \limits_{x \to 0} \bigg( \large\frac{ 2 \sin 4x \cos 2x}{2 \cos 4x \sin x} \bigg)$
$\qquad =\lim \limits_{x \to 0} \bigg( \large\frac{ \sin 4x \cos 2x}{ \cos 4x \sin x} \bigg)$
$\qquad =\lim \limits_{x \to 0} \bigg \{ \large\frac{ \sin 4x}{4x} \times \frac{x}{\sin x }$$ \times \cos 2x \times \large\frac{1}{\cos 4x}$$ \times 4 \bigg \}$
$\qquad = 4 \times \lim \limits _{4x \to 0} \large\frac{\sin 4x}{4x} $$\times \lim \limits _{x \to 0} \bigg(\large\frac{x}{\sin x} \bigg) $$\times \lim \limits_{2x \to 0} \cos 2x \times \lim \limits _{4x \to 0} \large\frac{1}{\cos 4x}$
$\qquad= (4 \times 1 \times 1 \times 1 \times \large\frac{1}{1} )$
Hence A is the correct answer.
answered Apr 22, 2014 by meena.p
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App