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MHTCET 26 Apr 2025 Shift 2 Paper

Question : 109 of 150

Marks: +1, -0

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

Solution:

Concept:

Use standard Taylor expansions near

x = 0

to replace each function by its leading term, then simplify the limit.

Explanation:

x \to 0

, both numerator and denominator tend to

0

Expand the numerator terms:

e^{x^{2}} = 1 + x^{2} + \cdots

\cos 3x = 1 - \frac{9x^{2}}{2} + \cdots

So,

e^{x^{2}} - \cos 3x = \left(1 + x^{2} + \cdots\right) - \left(1 - \frac{9x^{2}}{2} + \cdots\right) = \frac{11x^{2}}{2} + \cdots

Now expand the denominator terms:

\sin x = x + \cdots

\log(1 + 2x) = 2x + \cdots

Therefore,

\sin x \log(1 + 2x) = (x + \cdots)(2x + \cdots) = 2x^{2} + \cdots

Hence,

\lim_{x \to 0} \frac{e^{x^{2}} - \cos 3x}{\sin x \log(1 + 2x)} = \frac{\frac{11}{2}x^{2}}{2x^{2}} = \frac{11}{4}

Answer:

\frac{11}{4}

Option C.

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