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Chapter 4: Problem 28

Evaluate the following limits. $$\lim _{x \rightarrow 0} \frac{\sin x-x}{7 x^{3}}$$

Short Answer

Expert verified
Question: Evaluate the limit of the function as x approaches 0: $$\lim _{x \rightarrow 0} \frac{\sin x-x}{7 x^{3}}$$ Answer: The limit of the given function as x approaches 0 is: $$\lim_{x \rightarrow 0} \frac{\sin x-x}{7x^3} = -\frac{1}{42}$$

Step by step solution

01

Check for substitution

First, let's try to substitute x=0 into the limit. If you find no indeterminate form, it won't be necessary to apply L'H么pital's Rule. $$\frac{\sin 0-0}{7 (0)^{3}}=\frac{0}{0}$$ This limit is indeterminate, so let's continue with L'H么pital's Rule.
02

Apply L'H么pital's Rule once

Differentiate the numerator and denominator: Numerator: \(\frac{d}{dx} (\sin x - x) = \cos x - 1\) Denominator: \(\frac{d}{dx} (7x^3) = 21x^2\) Now, check the limit for the new function: $$\lim _{x \rightarrow 0} \frac{\cos x - 1}{21x^{2}}$$ Substitute 0 back into the function: $$\frac{\cos 0 - 1}{21(0)^{2}}=\frac{-1}{0}$$ This still doesn't give us a definite solution, so we must apply L'H么pital's Rule again.
03

Apply L'H么pital's Rule again

Differentiate the numerator and denominator again: Numerator: \(\frac{d}{dx} (\cos x - 1) = -\sin x\) Denominator: \(\frac{d}{dx} (21x^2) = 42x\) Evaluate the limit for the new function: $$\lim _{x \rightarrow 0} \frac{-\sin x}{42x}$$ Substitute 0 back into the function: $$\frac{-\sin 0}{42(0)}=\frac{0}{0}$$ This is also indeterminate, so let's apply L'H么pital's Rule one more time.
04

Apply L'H么pital's Rule for the third time

Differentiate the numerator and denominator again: Numerator: \(\frac{d}{dx}(-\sin x) = -\cos x\) Denominator: \(\frac{d}{dx} (42x) = 42\) Now, we have the following limit: $$\lim _{x \rightarrow 0} \frac{-\cos x}{42}$$
05

Final evaluation

Since the denominator is a constant, substitution will give a definite answer. Substitute x=0 into the limit: $$\frac{-\cos 0}{42}=\frac{-1}{42}$$ This is the final answer: $$\lim_{x \rightarrow 0} \frac{\sin x-x}{7x^3} = -\frac{1}{42}$$

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