Match List -I with List-II
| List-I | List-II | ||
| (A) | $\lim\limits_{x→0}\frac{1-cos2x}{x}$ | (I) | $\frac{5}{2}$ |
| (B) | $\lim\limits_{x→0}\frac{sin2x}{x}$ | (II) | 5 |
| (C) | $\lim\limits_{x→1}\frac{x^5-1}{x^2-1}$ | (III) | 2 |
| (D) | $\lim\limits_{x→0}\frac{(x+1)^5-1}{x}$ | (IV) | 0 |
Choose the correct answer from the options given below :
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A)-(IV), (B)-(III),(C)-(I),(D)-(II)
(A) $\lim\limits_{x→0}\frac{1-\cos 2x}{x}=\lim\limits_{x→0}\frac{2\sin^2x}{x}$
$=2×1×0=0$ (IV)
(B) $\lim\limits_{x→0}\frac{\sin 2x×2}{(x×2)}=1×2=2$ (III)
(C) $\lim\limits_{x→1}\frac{x^5-1}{x^2-1}=\frac{0}{0}form$
Using L'hopital's rule
$\lim\limits_{x→1}\frac{5x^4}{2x}=\frac{5}{2}$ (I)
(D) $\lim\limits_{x→0}\frac{(x+1)^5-1}{x}=\frac{0}{0}form$
Using L'hopital's rule
$\lim\limits_{x→0}\frac{5(x+1)^4}{1}=5$ (II)