Match List I with List II
| LIST I | LIST II | ||
| A. | $\lim\limits_{x→0}\frac{(1-cos2x)sin5x}{x^2sin3x}$ | I. | 18 |
| B. | $\lim\limits_{x→∞}\frac{(3x-5)(2x-7)}{(4x-9)(5x-3)}$ | II. | $\frac{10}{3}$ |
| C. | $\lim\limits_{x→0}\frac{2sin^23x}{x^2}$ | III. | $\frac{3}{4}$ |
| D. | $\lim\limits_{x→\frac{\pi}{4}}\frac{1-cos^3x}{2-cotx-cot^3x}$ | IV. | $\frac{3}{10}$ |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 3
A. $\displaystyle \lim_{x\to 0}\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x} =\lim_{x\to 0}\frac{\left(\frac{(2x)^2}{2}\right)\,(5x)}{x^2\,(3x)} =\frac{10}{3}$
B. $\displaystyle \lim_{x\to \infty}\frac{(3x-5)(2x-7)}{(4x-9)(5x-3)} =\lim_{x\to \infty}\frac{6x^2+\cdots}{20x^2+\cdots} =\frac{6}{20}=\frac{3}{10}$
C. $\displaystyle \lim_{x\to 0}\frac{2\sin^2 3x}{x^2} =\lim_{x\to 0}\,2\left(\frac{\sin 3x}{x}\right)^2 =2\cdot 3^2=18$
D. $\displaystyle \lim_{x\to \frac{\pi}{4}}\frac{1-\cos^3 x}{2-\cot x-\cot^3 x} =\frac{3}{4}$