Match List – I with List – II.

Choose the correct answer from the options given below:

A-IV, B-II, C-III, D-I

The correct answer is Option (2) → A-IV, B-II, C-III, D-I

(A) $\lim\limits_{x \rightarrow 0} \frac{1-\cos 2 x}{3 x^2}=\lim\limits_{x \rightarrow 0}\frac{2\sin 2x}{6x}=\lim\limits_{x \rightarrow 0}\frac{4\cos 2x}{6}$

$=\frac{4}{6}=\frac{2}{3}$

(B) $\lim\limits_{x \rightarrow 4} \frac{x^2-16}{x-4}=\lim\limits_{x \rightarrow 4}\frac{(x-4)(x+4)}{(x-4)}$

$⇒\lim\limits_{x \rightarrow 4}(x+4)=8$

(C) $\lim\limits_{x \rightarrow 0} \frac{\sin a x+4 x}{a x+\sin 4 x}=\lim\limits_{x \rightarrow 0}\frac{a\cos ax+4}{a+4\cos 4x}$

$=\frac{a+4}{a+4}=1$

(D) $\lim\limits_{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}$

$\lim\limits_{z \to 1} \frac{z^{\frac{1}{3}} - 1}{z^{\frac{1}{6}} - 1} = \lim\limits_{z \to 1} \frac{\frac{1}{3} z^{-\frac{2}{3}}}{\frac{1}{6} z^{-\frac{5}{6}}}$

$\frac{\frac{1}{3} z^{-\frac{2}{3}}}{\frac{1}{6} z^{-\frac{5}{6}}} = \frac{1}{3} \times \frac{6}{1} \times z^{\frac{5}{6} - \frac{2}{3}} = 2 \times z^{\frac{5}{6} - \frac{4}{6}} = 2 \times z^{\frac{1}{6}}$

$\lim\limits_{z \to 1} 2 \times z^{\frac{1}{6}} = 2 \times 1^{\frac{1}{6}} = 2 \times 1 = 2$