lim _{x rightarrow 0} frac{e^{x^2}-cos 3 x}{s | vedclass

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Question

(C) We evaluate the limit using the standard series expansions for $x \rightarrow 0$: $e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \dots$ $\cos 3x = 1 - \frac

Vedclass · Accepted Answer

$\frac{11}{4}$

Vedclass · Answer

(C) We evaluate the limit using the standard series expansions for $x \rightarrow 0$: $e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \dots$ $\cos 3x = 1 - \frac{(3x)^2}{2} + \dots = 1 - \frac{9x^2}{2} + \dots$ $\sin x = x + \dots$ $\log(1+2x) = 2x - \frac{(2x)^2}{2} + \dots = 2x - 2x^2 + \dots$ Substituting these into the expression: $\lim _{x \rightarrow 0} \frac{(1 + x^2 + \dots) - (1 - \frac{9x^2}{2} + \dots)}{(x + \dots)(2x - \dots)} = \lim _{x \rightarrow 0} \frac{x^2 + \frac{9x^2}{2}}{2x^2} = \lim _{x \rightarrow 0} \frac{\frac{11x^2}{2}}{2x^2} = \frac{11}{4}$.

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

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