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MHTCET 18 Apr 2026 Shift 2 Paper

Question : 108 of 150

Marks: +1, -0

\lim _{x \rightarrow k} \frac{x^{3}-k^{3}}{x^{2}-k^{2}}=\lim _{x \rightarrow 0} \frac{1-\cos (2 x)}{x \sin x}

k

is . . .

Solution:

Concept:

Use standard limit formulae and algebraic factorization to equate both sides.

Explanation:

First, evaluate the limit on the right-hand side.

Use

1-\cos 2x = 2\sin^2 x

Then,

\lim_{x \to 0} \frac{1-\cos 2x}{x\sin x} = \lim_{x \to 0} \frac{2\sin^2 x}{x\sin x} = \lim_{x \to 0} 2 \cdot \frac{\sin x}{x}

Since

\lim_{x \to 0} \frac{\sin x}{x} = 1

, the right-hand side equals

2

Now simplify the left-hand side.

Factorize:

x^3 - k^3 = (x-k)(x^2 + xk + k^2)

and

x^2 - k^2 = (x-k)(x+k)

Cancel

(x-k)

to get

\lim_{x \to k} \frac{x^2 + xk + k^2}{x+k}

Substitute

x = k

\frac{3k^2}{2k} = \frac{3k}{2}

Equate both limits:

\frac{3k}{2} = 2

Thus,

3k = 4

, so

k = \frac{4}{3}

Answer:

k = \frac{4}{3}

, which is option A.

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