/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 24 Evaluate the following limits. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Evaluate the following limits. $$\lim _{x \rightarrow 0} \frac{\sin ^{2} 3 x}{x^{2}}$$

Short Answer

Expert verified
$$\lim_{x\rightarrow 0} \frac{\sin^{2}(3x)}{x^2}$$ Answer: The limit of the given expression as x approaches 0 is 18.

Step by step solution

01

Simplify the expression

Rewrite the expression as: $$\lim_{x\rightarrow 0} \frac{\sin^{2}(3x)}{x^2} = \lim_{x\rightarrow 0} \frac{(\sin(3x))^2}{x^2}$$
02

Apply L'Hôpital's Rule

Since the limit is an indeterminate form of type \(\frac{0}{0}\) as \(x\rightarrow 0\), we can use L'Hôpital's Rule. Apply it twice, as we have a squared function: $$\lim_{x\rightarrow 0} \frac{(\sin(3x))^2}{x^2} = \lim_{x\rightarrow 0} \frac{2\sin(3x) \cos(3x) \cdot 3}{2x}$$ After simplifying this expression: $$= \lim_{x\rightarrow 0} \frac{6\sin(3x)\cos(3x)}{x}$$
03

Apply L'Hôpital's Rule again

As the limit is still an indeterminate form of type \(\frac{0}{0}\) as \(x\rightarrow 0\), we can apply L'Hôpital's Rule once more: $$\lim_{x\rightarrow 0} \frac{6\sin(3x)\cos(3x)}{x} = \lim_{x\rightarrow 0} \frac{6(\cos^{2}(3x)-\sin^{2}(3x))\cdot 3}{1}$$
04

Evaluate the limit

Now the limit is no longer indeterminate, so we can directly substitute x=0 into the expression: $$\lim_{x\rightarrow 0} 18(\cos^2(3x) - \sin^2(3x)) = 18(\cos^2(0)-\sin^2(0)) = 18(1^2 - 0^2) = 18$$ Hence, the limit of the given expression as x approaches 0 is 18.

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