Question

The value of \( \lim_{x\rightarrow0}\frac{e^{2x^{2}}-\cos 2x}{x^{2}} = \)

• A. 2
• B. \( \frac{5}{2} \)
• C. 4
• D. 6

Hint:

L'Hôpital's rule is an alternative strategy for evaluating limits of the form \( \frac{0}{0} \). Differentiating the numerator and denominator separately will quickly clear the indeterminate state.

Answer 1

The Correct Option is C

Concept: Substituting \( x = 0 \) results in the indeterminate form \( \frac{1-1}{0} = \frac{0}{0} \). We can solve this quickly using Taylor series expansions around \( x = 0 \):

• \( e^t = 1 + t + \frac{t^2}{2!} + \dots \implies e^{2x^2} = 1 + 2x^2 + \frac{(2x^2)^2}{2} + \dots \)

• \( \cos t = 1 - \frac{t^2}{2!} + \dots \implies \cos 2x = 1 - \frac{(2x)^2}{2} + \dots = 1 - 2x^2 + \dots \)

Step 1: Expanding the terms in the numerator.
\[ e^{2x^2} - \cos 2x = (1 + 2x^2 + O(x^4)) - (1 - 2x^2 + O(x^4)) \] Subtracting the expansions: \[ = (1 - 1) + (2x^2 - (-2x^2)) + O(x^4) = 4x^2 + O(x^4) \] Let us re-verify second derivative scales using L'Hôpital's rule: \[ \text{First derivative} = \frac{4xe^{2x^2} + 2\sin 2x}{2x} = \frac{4e^{2x^2} + 4\cos 2x}{2} \rightarrow \frac{4+4}{2} = 4 \]