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MHTCET 19 Apr 2026 Shift 1 Paper

Question : 111 of 150

Marks: +1, -0

The value of

\lim _{x \rightarrow 0}\left(\frac{8}{x^{8}}\right)\left[1-\cos \frac{x^{2}}{2}-\cos \frac{x^{2}}{4}+\cos \frac{x^{2}}{2} \cdot \cos \frac{x^{2}}{4}\right]

is equal to ...

Solution:

Concept:

Factorise the four-term trigonometric expression into a product of two simpler terms and then apply the standard limit

\lim_{t \to 0} \frac{1-\cos t}{t^2} = \frac{1}{2}

Explanation:

The bracket can be factorised as

1-\cos \frac{x^{2}}{2}-\cos \frac{x^{2}}{4}+\cos \frac{x^{2}}{2}\cos \frac{x^{2}}{4} = \left(1-\cos \frac{x^{2}}{2}\right)\left(1-\cos \frac{x^{2}}{4}\right) .

So the limit becomes

\lim_{x \to 0} \frac{8}{x^{8}}\left(1-\cos \frac{x^{2}}{2}\right)\left(1-\cos \frac{x^{2}}{4}\right) .

Using the standard result

\lim_{t \to 0} \frac{1-\cos t}{t^{2}} = \frac{1}{2}

, we get the approximations

1-\cos \frac{x^{2}}{2} \approx \frac{1}{2}\left(\frac{x^{2}}{2}\right)^{2} = \frac{x^{4}}{8},

and

1-\cos \frac{x^{2}}{4} \approx \frac{1}{2}\left(\frac{x^{2}}{4}\right)^{2} = \frac{x^{4}}{32}.

Substituting these into the limit,

\lim_{x \to 0} \frac{8}{x^{8}} \cdot \frac{x^{4}}{8} \cdot \frac{x^{4}}{32} = \frac{8}{256} = \frac{1}{32}.

Thus, the correct value of the limit is

\frac{1}{32}

Answer:

\frac{1}{32}

Option B.

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