Evaluate the limit: lim _{x rightarrow 0} fra | vedclass

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(B) Let $L = \lim _{x ightarrow 0} \frac{	an ^2(\pi \sec ^4 x)}{\pi^2 x^4}$.Since $\sec(0) = 1$,the expression takes the form $\frac{	an^2(\pi)}{0

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Let $L = \lim _{x ightarrow 0} \frac{	an ^2(\pi \sec ^4 x)}{\pi^2 x^4}$.Since $\sec(0) = 1$,the expression takes the form $\frac{	an^2(\pi)}{0} = \frac{0}{0}$.We use the property $	an(\pi \sec^4 x) = 	an(\pi \sec^4 x - \pi) = 	an(\pi(\sec^4 x - 1))$.Recall that $\sec^4 x - 1 = (\sec^2 x - 1)(\sec^2 x + 1) = 	an^2 x (\sec^2 x + 1)$.As $x ightarrow 0$,$	an x \approx x$ and $\sec^2 x \approx 1$,so $\sec^4 x - 1 \approx x^2(1+1) = 2x^2$.Thus,$	an(\pi \sec^4 x) \approx 	an(2\pi x^2) \approx 2\pi x^2$.Substituting this into the limit:$L = \lim _{x ightarrow 0} \frac{(2\pi x^2)^2}{\pi^2 x^4} = \lim _{x ightarrow 0} \frac{4\pi^2 x^4}{\pi^2 x^4} = 4$.

Evaluate the limit: $\lim _{x \rightarrow 0} \frac{\tan ^2(\pi \sec ^4 x)}{\pi^2 x^4}$

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