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

Step 1: Check the form.
Putting $x=0$ gives $\frac{e^0-\cos0}{0}=\frac{1-1}{0}=\frac{0}{0}$, which is indeterminate, so we expand each term.
Step 2: Expand $e^{2x^2}$.
Using $e^t=1+t+\frac{t^2}{2}+\dots$ with $t=2x^2$: \[ e^{2x^2}=1+2x^2+\dots \]
Step 3: Expand $\cos 2x$.
Using $\cos t=1-\frac{t^2}{2}+\dots$ with $t=2x$: \[ \cos 2x=1-\frac{(2x)^2}{2}+\dots=1-2x^2+\dots \]
Step 4: Subtract in the numerator.
\[ e^{2x^2}-\cos 2x=(1+2x^2)-(1-2x^2)+\dots=4x^2+\dots \]
Step 5: Divide by $x^2$.
\[ \frac{4x^2+\dots}{x^2}=4+(\text{terms that vanish as }x\to0) \]
Step 6: Take the limit.
\[ \lim_{x\to0}\frac{e^{2x^2}-\cos 2x}{x^2}=4 \] \[ \boxed{4} \]