$\lim _{x \rightarrow 0}\left(\left(\frac{1-\cos ^{2}(3 x)}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)$ is equal to

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Limits

lim_{x \to 0} ((\frac{1 - c o s ^{2} ( 3 x )}{c o s ^{3} ( 4 x )}) (\frac{s i n ^{3} ( 4 x )}{( l o g _{e} ( 2 x + 1 ) ) ^{5}}))

is equal to

Answer: (B)

Rewrite

1 - cos^{2} (3 x) = sin^{2} (3 x)

.

The limit becomes:

lim_{x \to 0} \frac{s i n ^{2} ( 3 x )}{c o s ^{3} ( 4 x )} \cdot \frac{s i n ^{3} ( 4 x )}{( l o g _{e} ( 2 x + 1 ) ) ^{5}}

Using standard limits

\frac{s i n ( a x )}{a x} \to 1

and

\frac{l o g ( 1 + b x )}{b x} \to 1

= lim_{x \to 0} \frac{( 3 x ) ^{2}}{1} \cdot \frac{( 4 x ) ^{3}}{( 2 x ) ^{5}} = \frac{9 x ^{2} \cdot 64 x ^{3}}{32 x ^{5}} = \frac{576 x ^{5}}{32 x ^{5}} = 18

JEE Main · 2023 · 08 Apr 2023 · Shift 1 · Single correct option

Let $f : R \to (0, \infty)$ be a twice differentiable function such that $f (3) = 18, f^{'} (3) = 0$ and $f^{''} (3) = 4$ .
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