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Limits Continuity and Differentiability question

2023 · 8 Apr · Shift 1 · Q25

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Limits Continuity and Differentiability
2023 · 8 Apr · Shift 1 · Q25

Limits Continuity and Differentiability question

2023 · 8 Apr · Shift 1 · Q25

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1

\lim_{x \rightarrow 0}\left(\left(\frac{\left(1-\cos ^{2}(3 x)\right.}{\cos ^{3}(4 x)}\right)\left(\frac{\sin ^{3}(4 x)}{\left(\log _{e}(2 x+1)\right)^{5}}\right)\right)

is equal to

\underline{\hspace{2cm}}

.

A
15
B
18
C
9
D
24

View written solutionFree

Correct answer: B

Rewrite the expression

We need to evaluate

\lim_{x\to 0}\left(\frac{1-\cos^2(3x)}{\cos^3(4x)}\cdot \frac{\sin^3(4x)}{(\ln(1+2x))^5}\right).

Use the identity

$1-\cos^2(3x)=\sin^2(3x).$

So the limit becomes

\lim_{x\to 0}\frac{\sin^2(3x)\sin^3(4x)}{\cos^3(4x)\, (\ln(1+2x))^5}.

Use standard limits near $x=0$

As $x\to 0$ ,

$\sin(ax) \sim ax, \qquad \ln(1+2x) \sim 2x, \qquad \cos(4x)\to 1.$

Hence,

\sin^2(3x) \sim (3x)^2 = 9x^2,

\sin^3(4x) \sim (4x)^3 = 64x^3,

(\ln(1+2x))^5 \sim (2x)^5 = 32x^5,

and

\cos^3(4x) \to 1.

Substitute these approximations

Therefore,

\lim_{x\to 0}\frac{\sin^2(3x)\sin^3(4x)}{\cos^3(4x)\, (\ln(1+2x))^5} = \lim_{x\to 0}\frac{9x^2\cdot 64x^3}{1\cdot 32x^5}.

Now simplify:

\frac{9\cdot 64x^5}{32x^5} = 9\cdot 2 = 18.

So the limit is

$\boxed{18}.$

Check options

A: $15$ ❌
B: $18$ ✅
C: $9$ ❌
D: $24$ ❌

Therefore, the correct option is B.

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