Week 7 Activity

Section 7.4 Week 7 Activity

Subsection 7.4.1 Chain Rule

Activity 7.4.1.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \sin (x^2 + 1) \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = \sin u\text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = x^2 +1\text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \sin (x^2 + 1) = \frac{d}{du} \sin u \Bigg|_{u=x^2+1} \frac{d}{dx} (x^2+1) \end{equation*}

The derivative of sine is a known derivative, and linearity and the power rule give the derivative of the polynomial term.

\begin{equation*} = \cos u \Bigg|_{u=x^2+1} (2x+0) \end{equation*}

After I’ve done the derivative, I replace \(u\) as the evaluation bar reminds me to do. I’ll also change the order of the terms, since convention is to usually write the polynomial term before the trigonometric term.

\begin{equation*} = 2x \cos (x^2+1) \end{equation*}

Activity 7.4.2.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \ln (\ln x) \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = \ln u \text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = \ln x\text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \ln (\ln x) = \frac{d}{du} \ln u \Bigg|_{u = \ln x} \frac{d}{dx} \ln x \end{equation*}

The deriavtive of the logarithm is a known derivative.

\begin{equation*} = \frac{1}{u} \Bigg|_{u = \ln x} \frac{1}{x} = \frac{1}{x \ln x} \end{equation*}

Activity 7.4.3.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \frac{1}{1+e^x} \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = \frac{1}{u} = u^{-1}\text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = 1 + e^x \text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \frac{1}{1+e^x} = \frac{d}{du} u^{-1} \Bigg|_{u = 1 + e^x} \frac{d}{dx} (1+e^x) \end{equation*}

The first derivative is a power rule derivative. Linearity and the known derivative of the exponential function give me the second derivative.

\begin{equation*} = -u^{-2} \Bigg|_{u = 1 + e^x} (e^x) \end{equation*}

Then I replace \(u\text{.}\)

\begin{equation*} = \frac{-1}{(1+e^x)^2} (e^x) = \frac{-e^x}{(1+e^x)^2} \end{equation*}

Activity 7.4.4.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \sqrt[4]{x^2+4x+2} \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = \sqrt[4]{u} = u^\frac{1}{4} \text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = x^2 + 4x + 2\text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \sqrt[4]{x^2+4x+2} = \frac{d}{du} u^\frac{1}{4} \Bigg|_{u = x^2 + 4x + 2} \frac{d}{dx} (x^2 + 4x + 2) \end{equation*}

I use power rule for the first derivative (which is why I wrote the fourth root as a fractional exponent). The polynomial derivative is solved by linearity and the power rule.

\begin{equation*} = \frac{1}{4} u^\frac{-3}{4} \Bigg|_{u = x^2 + 4x + 2} (2x + 4) \end{equation*}

Then I just have to do the replacement. I’ll also write the fractional exponent as a root again.

\begin{equation*} = \frac{1}{4\sqrt[4]{u^3}} (2x + 4) = \frac{2x+4}{4 \sqrt[4]{x^2+4x+2)^3}} \end{equation*}

Activity 7.4.5.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \cos \left( e^{-4x} \right) \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = \cos (u) \text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = e^{-4x} \text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \cos \left( e^{-4x} \right) = \frac{d}{du} \cos u \Bigg|_{u = e^{-4x}} \frac{d}{dx} e^{-4x} \end{equation*}

The derivative of cosine is known. However, this second derivative is still a composition, so I need to use chain rule again. To avoid confusion of variables, I’ll use a new temporary variable \(v\text{.}\) The outside is \(f(v) = e^v\) and the inside is \(g(x) = -4x \text{.}\) I apply the chain rule again using this decomposition.

\begin{equation*} = -\sin u \Bigg|_{u = e^{-4x}} \frac{d}{dv} e^v \Bigg|_{v = -4x} \frac{d}{dx} (-4x) \end{equation*}

Then I proceed with the known derivatives, replacing \(u\) and \(v\) when I have finished the derivatives.

\begin{align*} \amp = -\sin \left( e^{-4x} \right) e^v \Bigg|_{v = -4x} (-4) \\ \amp = -\sin \left( e^{-4x} \right) e^{-4x} (-4) = 4 e^{-4x} \sin \left( e^{-4x} \right) \end{align*}

Note that I could have chosen a different decomposition. I made the choice I made to make the outside function as simple as possible. When you have a complicated composition of several functions, work from the outside in. Make the outside function easily differentiable, and apply more rules to the inside function if necessary.

Activity 7.4.6.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \sqrt{\sin(x^2+1)} \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = \sqrt{u} = u^\frac{1}{2}\text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = \sin (x^2+1)\text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \sqrt{\sin(x^2+1)} = \frac{d}{du} u^\frac{1}{2} \Bigg|_{u = \sin (x^2 + 1)} \frac{d}{dx} \sin (x^2+1) \end{equation*}

I need to use the chain rule again for the second derivative. The outside is \(\sin v\) and the inside is \((x^2 +1)\text{.}\)

\begin{equation*} = \frac{1}{2} u^\frac{-1}{2} \Bigg|_{u = \sin (x^2 + 1)} \frac{d}{dv} \sin v \Bigg|_{v = x^2 +1} \frac{d}{dx} x^2+1 \end{equation*}

Now I just need to finish the individual derivative and re-arrange the answer into a reasonable form.

\begin{align*} \amp = \frac{1}{2\sqrt{u}} \Bigg|_{u = \sin (x^2 + 1)} \cos v \Bigg|_{v = x^2 +1} (2x)\\ \amp = \frac{1}{2\sqrt{\sin(x^2+1))}} \cos (x^2+1) (2x)\\ \amp = \frac{x\cos(x^2+1)}{\sqrt{\sin(x^2+1))}} \end{align*}

Subsection 7.4.2 Higher Derivatives

Activity 7.4.7.

Calculate this derivative.

\begin{equation*} \frac{d^2}{dx^2} 5x^4 - 7x^3 + 15x + 1 \end{equation*}

Solution.

I need to take the derivative twice. Both times I am taking polynomial derivatives, so I am using linearity and the power rule for the derivatives.

\begin{align*} \frac{d^2}{dx^2} 5x^4 - 7x^3 + 15x + 1 \amp = \frac{d}{dx} \frac{d}{dx} 5x^4 - 7x^3 + 15x + 1 \\ \amp = \frac{d}{dx} 20x^3 - 21x^2 + 15 \\ \amp = 60x^2 - 42x \end{align*}

Activity 7.4.8.

Calculate this derivative.

\begin{equation*} \frac{d^3}{dx^3} e^{4x} \end{equation*}

Solution.

I need to take the derivative three times. Each time, this is a chain rule derivative, with outside function \(e^u\) and inside function \(4x\text{.}\) Each time the derivative of the inside produces a constant 4, which I can factor out of the remaining steps due to linearity.

\begin{align*} \frac{d^3}{dx^3} e^{4x} \amp = \frac{d}{dx} \frac{d}{dx} \frac{d}{du} e^u \Bigg|_{u = 4x} \frac{d}{dx} 4x\\ \amp = \frac{d}{dx} \frac{d}{dx} e^u \Bigg|_{u = 4x} 4\\ \amp = 4 \frac{d}{dx} \frac{d}{dx} e^{4x} \\ \amp = 4 \frac{d}{dx} \frac{d}{du} e^{u} \Bigg|_{u=4x} \frac{d}{dx} 4x \\ \amp = 4 \frac{d}{dx} e^{u} \Bigg|_{u=4x} 4 \\ \amp = 16 \frac{d}{dx} e^{4x}\\ \amp = 16 \frac{d}{du} e^{u} \Bigg|_{u=4x} \frac{d}{dx} 4x \\ \amp = 16 e^{u} \Bigg|_{u=4x} 4 = 64 e^{4x} \end{align*}

Activity 7.4.9.

Calculate this derivative.

\begin{equation*} \frac{d^4}{dx^4} \sin x - \cos x \end{equation*}

Solution.

I have four derivatives to take. In each step, I use linearity and the known derivatives of sine and cosine.

\begin{align*} \frac{d^4}{dx^4} \sin x - \cos x \amp = \frac{d}{dx} \frac{d}{dx} \frac{d}{dx} \frac{d}{dx} \sin x - \cos x \\ \amp = \frac{d}{dx} \frac{d}{dx} \frac{d}{dx} \cos x + \sin x \\ \amp = \frac{d}{dx} \frac{d}{dx} \frac{d}{dx} \cos x + \sin x \\ \amp = \frac{d}{dx} \frac{d}{dx} -\sin x + \cos x \\ \amp = \frac{d}{dx} -\cos x - \sin x \\ \amp = \sin x - \cos x \end{align*}

Notice that I return to the original expression. That is as expected, since the fourth derivative of both sine and cosine is the same as whichever I started with. This subtraction of one from the other also has the property of being preserved under four derivatives, since the derivative respects subtraction.

Subsection 7.4.3 L’Hôpital’s Rule

Activity 7.4.10.

Check that this limit is an indeterminate form of type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty}\text{.}\) Use L’Hôpital’s Rule to calculate the limit.

\begin{equation*} \lim_{x \rightarrow 0} \frac{\cos x - 1}{x} \end{equation*}

Solution.

Since \(\cos 0 = 1\text{,}\) the numerator approaches 0. The denominator also approaches 0, so this is a indeterminate form of type \(\frac{0}{0}\text{.}\) I apply L’Hôpital’s rule. After applying the rule the resulting limit can be directly evaluted.

\begin{align*} \lim_{x \rightarrow 0} \frac{\cos x - 1}{x} \amp = \lim_{x \rightarrow 0} \frac{ \frac{d}{dx} (\cos x - 1)}{\frac{d}{dx} x} \\ \amp = \lim_{x \rightarrow 0} \frac{ -\sin x - 0}{1} = \frac{0}{1} = 0 \end{align*}

Activity 7.4.11.

Check that this limit is an indeterminate form of type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty}\text{.}\) Use L’Hôpital’s Rule to calculate the limit. (Don’t solve the limit by factoring.)

\begin{equation*} \lim_{x \rightarrow 4} \frac{x^2 - 7x + 12}{x^2 + x - 20} \end{equation*}

Solution.

The limit of the numerator as \(x \rightarrow 4\) is \(16 - 28 + 12 = 0\text{.}\) The limit of the denominator as \(x \rightarrow 4\) is \(16 + 4 - 20 = 0\text{.}\) This is an indeterminate form of type \(\frac{0}{0}\) and I can apply L’Hôpital’s rule. After applying the rule, the resulting limit can be directly evaluated.

\begin{align*} \lim_{x \rightarrow 4} \frac{x^2 - 7x + 12}{x^2 + x - 20} \amp = \lim_{x \rightarrow 4} \frac{\frac{d}{dx} (x^2 - 7x + 12)}{\frac{d}{dx} (x^2 + x + 20)}\\ \amp = \lim_{x \rightarrow 4} \frac{2x-7}{2x+1} = \frac{8-7}{8+1} = \frac{1}{9} \end{align*}

Activity 7.4.12.

Check that this limit is an indeterminate form of type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty}\text{.}\) Use L’Hôpital’s Rule to calculate the limit.

\begin{equation*} \lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}} \end{equation*}

Solution.

Both the functions have asymptotes as I approach \(0^+\text{,}\) so this is an indeterminate form of type \(\frac{\infty}{\infty}\text{.}\) I can apply L’Hôpital’s rule.

\begin{align*} \lim_{x \rightarrow 0^+} \frac{\ln x}{\frac{1}{x}} \amp = \lim_{x \rightarrow 0^+} \frac{\frac{d}{dx} \ln x}{\frac{d}{dx} x^{-1}} \\ \amp = \lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{\frac{-1}{x^2}} \end{align*}

Now I have to simplify the nested fraction. I’ll remind you of the general rule for simplifying nested tractions.

\begin{equation*} \frac{\frac{a}{b}}{\frac{c}{d}} = \frac{ad}{bc} \end{equation*}

I apply this rule to the previous expression to finish the limit. After canceling \(x\) from numerator and denominator, the resulting limit can be directly evaluated.

\begin{equation*} \lim_{x \rightarrow 0^+} \frac{\frac{1}{x}}{\frac{-1}{x^2}} = \lim_{x \rightarrow 0^+} \frac{x^2}{-x} = \lim_{x \rightarrow 0^+} -x = 0 \end{equation*}

Activity 7.4.13.

Check that this limit is an indeterminate form of type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty}\text{.}\) Use L’Hôpital’s Rule to calculate the limit. (Don’t use asymptotic analysis.)

\begin{equation*} \lim_{x \rightarrow \infty} \frac{x^2\ln x}{x^3+7} \end{equation*}

Solution.

Both numerator and denominator are growing functions without bound, so this is an indeterminate form of type \(\frac{\infty}{\infty}\text{.}\) I can apply L’Hôpital’s rule.

\begin{align*} \lim_{x \rightarrow \infty} \frac{x^2\ln x}{x^3+7} \amp = \lim_{x \rightarrow \infty} \frac{\frac{d}{dx} x^2 \ln x}{\frac{d}{dx} x^3+7} \\ \amp = \lim_{x \rightarrow \infty} \frac{2x \ln x + x^2 \frac{1}{x}}{3x^2+0}\\ \amp = \lim_{x \rightarrow \infty} \frac{2x \ln x + x}{3x^2+0}\\ \amp = \lim_{x \rightarrow \infty} \frac{x (2\ln x + 1)}{3x^2}\\ \amp = \lim_{x \rightarrow \infty} \frac{2\ln x + 1}{3x} \end{align*}

This cannot be directly evaluated. However, I can use L’Hôpital’s rule again. Both functions are still growing to \(\infty\text{,}\) so the rule still applies.

\begin{align*} \amp = \lim_{x \rightarrow \infty} \frac{\frac{d}{dx} 2\ln x + 1}{ \frac{d}{dx} 3x}\\ \amp = \lim_{x \rightarrow \infty} \frac{\frac{2}{x}}{3}\\ \amp = \lim_{x \rightarrow \infty} \frac{2}{3x} = 0 \end{align*}

This final limit can be directly determined: as \(x\) gets very large, its reciprocal gets very small and the limit goes to 0.

Subsection 7.4.4 Conceptual Review Questions

What are the inside and outside functions in a composition?
Why do compositions need their own unique derivative rule? Why can’t we use the previous rules (power, linearity, product or quotient) for compositions?
What does it meant to take a second or third derivative?
What does L’Hôpital’s rule have to do with asymptotic analysis?

Subsection 7.4.5 Extra Practice

Activity 7.4.14.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \ln (1+x) \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = \ln u\text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = 1 + x\text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \ln (1+x) = \frac{d}{du} \ln u \Bigg|_{u=1+x} \frac{d}{dx} (1+x) \end{equation*}

The derivative of the logarithm is a known derivative, and linearity determines the derivaitve of the inside.

\begin{equation*} = \frac{1}{u} \Bigg|_{u=1+x} (1) \end{equation*}

After I’ve done the derivative, I replace \(u\) as the evaluation bar reminds me to do.

\begin{equation*} = \frac{1}{1+x} \end{equation*}

Activity 7.4.15.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} e^{x^2+1} \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = e^u \text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = x^2 + 1\text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} e^{x^2+1} = \frac{d}{du} e^u \Bigg|_{u=x^2 + 1} \frac{d}{dx} (x^2 + 1) \end{equation*}

The derivative of the exponential is a known derivative, and linearity determines the derivaitve of the inside.

\begin{equation*} = e^u \Bigg|_{u=x^2+1} (2x) \end{equation*}

After I’ve done the derivative, I replace \(u\) as the evaluation bar reminds me to do.

\begin{equation*} = 2x e^{x^2+1} \end{equation*}

Activity 7.4.16.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \arctan (x^2-1) \end{equation*}

Solution.

This is a composition, so I use the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = \arctan u\text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = x^2 - 1\text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \arctan (x^2 - 1) = \frac{d}{du} \arctan u \Bigg|_{u=x^2 - 1} \frac{d}{dx} (x^2 - 1) \end{equation*}

The derivative of the arctangent is a known derivative, and linearity determines the derivaitve of the inside.

\begin{equation*} = \frac{1}{1 + u^2} \Bigg|_{u=x^2 - 1} (2x) \end{equation*}

After I’ve done the derivative, I replace \(u\) as the evaluation bar reminds me to do.

\begin{equation*} = \frac{2x}{1 + (x^2 - 1)^2} \end{equation*}

Activity 7.4.17.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} \ln (\arctan \sqrt{x}) \end{equation*}

Solution.

This is a double composition, so I use the chain rule twice. For the first chain rule, I’ll label the pieces: I’ll call the outside function \(f(u) = \ln u\text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = \arctan \sqrt{x} \text{.}\) Then I proceed with the chain rule.

\begin{equation*} \frac{d}{dx} \ln (\arctan \sqrt{x}) = \frac{d}{du} \ln u \Bigg|_{u = \arctan \sqrt{x}} \frac{d}{dx} \arctan \sqrt{x} \end{equation*}

The derivative of the logarithm is a known derivative. I need a second chain rule for the derivative of the inside. I’ll call the outside function \(g(v) = \arctan v \text{,}\) using the temporary variable \(v\text{.}\) The inside function is \(g(x) = \sqrt{x} \text{.}\) Then I proceed with the chain rule.

\begin{equation*} = \frac{1}{u} \Bigg|_{u = \arctan \sqrt{x}} \frac{d}{dv} \arctan v \Bigg|_{v = \sqrt{x}} \frac{d}{dx} \sqrt{x} \end{equation*}

The derivative of arctangent is known. The derivtive of the square can be done with the power rule, as I’ve done several times already in these notes.

\begin{equation*} = \frac{1}{u} \Bigg|_{u = \arctan \sqrt{x}} \frac{1}{1+v^2} \Bigg|_{v = \sqrt{x}} \frac{1}{2\sqrt{x}} \end{equation*}

After I’ve done the derivatives, I replace \(u\) and \(v\) as the evaluation bar reminds me to do.

\begin{equation*} = \frac{1}{(\arctan \sqrt{x})(1+\sqrt{x}^2)(2 \sqrt{x})} = \frac{1}{2(1+x)(\sqrt{x})(\arctan \sqrt{x})} \end{equation*}

Activity 7.4.18.

Calculate this derivative.

\begin{equation*} \frac{d}{dx} (\sin x)^2 + 3 (\sin x) \end{equation*}

Solution.

First I need linearity.

\begin{equation*} = \frac{d}{dx} (\sin x)^2 + 3 (\sin x) = \frac{d}{dx} (\sin x)^2 + 3 \frac{d}{dx} \sin x \end{equation*}

The derivative of the second term is the know derivative of sine. For the first term, I need the chain rule. I’ll label the pieces: I’ll call the outside function \(f(u) = u^2\text{,}\) using the temporary variable \(u\text{.}\) The inside function is \(g(x) = \sin x \text{.}\) Then I proceed with the chain rule (I’ll keep the second term in the linearity step around, of course.)

\begin{equation*} = \frac{d}{du} u^2 \Bigg|_{u = \sin x} \frac{d}{dx} \sin x + 3 \cos x \end{equation*}

The derivative of outside is the power rule and the derivative of the inside is the known derivative of sine.

\begin{equation*} = 2u \Bigg|_{u = \sin x} (cos x) + 3 \cos x \end{equation*}

After I’ve done the derivative, I replace \(u\) as the evaluation bar reminds me to do.

\begin{equation*} = 2 \sin x \cos x + 3 \cos x \end{equation*}

Activity 7.4.19.

Check that this limit is an indeterminate form of type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty}\text{.}\) Use L’Hôpital’s Rule to calculate the limit. (Don’t use asymptotic analysis.)

\begin{equation*} \lim_{x \rightarrow \infty} \frac{\ln x}{\sqrt[9]{x}} \end{equation*}

Solution.

Both numerator and denominator are fucntions that grow without bound, so this is a indeterminate form of type \(\frac{\infty}{\infty0}\text{.}\) I apply L’Hôpital’s rule. After applying the rule, the powers of \(x\) in the nested fraction can be simplified down (using the rules of exponents) into a limit that can be directly evaluated.

\begin{align*} \lim_{x \rightarrow \infty} \frac{\ln x}{\sqrt[9]{x}} \amp = \lim_{x \rightarrow \infty} \frac{ \frac{d}{dx} \ln x}{ \frac{d}{dx} \sqrt[9]{x} }\\ \amp = \lim_{x \rightarrow \infty} \frac{ \frac{1}{x}}{ \frac{1}{9\sqrt[9]{x^8}}} = \lim_{x \rightarrow \infty} \frac{x^{\frac{8}{9}}}{x} = \lim_{x \rightarrow \infty} \frac{1}{x^{\frac{1}{9}}} = 0 \end{align*}

In the final limit, only the denominator is growing, to the limit must converge to zero.

Activity 7.4.20.

Check that this limit is an indeterminate form of type \(\frac{0}{0}\) or type \(\frac{\infty}{\infty}\text{.}\) Use L’Hôpital’s Rule to calculate the limit.

\begin{equation*} \lim_{x \rightarrow 0} \frac{\sin x}{x^3-x} \end{equation*}

Solution.

Since \(\sin 0 = 0\text{,}\) the numerator approaches 0. The denominator also approaches \(0^3 - 0 = 0\text{,}\) so this is a indeterminate form of type \(\frac{0}{0}\text{.}\) I apply L’Hôpital’s rule. After applying the rule the resulting limit can be directly evaluted.

\begin{align*} \lim_{x \rightarrow 0} \frac{\sin x}{x^3 - x} \amp = \lim_{x \rightarrow 0} \frac{ \frac{d}{dx} \sin x}{\frac{d}{dx} (x^3 -x)}\\ \amp = \lim_{x \rightarrow 0} \frac{\cos x}{3x^2 - 1} = \frac{1}{0-1} = -1 \end{align*}

Prev Top Next