Question

t 36–42. calculator limits estimate the following limits using graphs or tables. 36. $limlimits_{h \to 0} \frac{(1 + 2h)^{1/h}}{2e^{2+h}}$ 37. $limlimits_{x \to pi/2} \frac{cot 3x}{cos x}$

Explanation:

Problem 36:

Step 1: Analyze the numerator limit

We know that the limit formula $\lim_{t \to 0}(1 + at)^{\frac{1}{t}}=e^{a}$. For the numerator $\lim_{h \to 0}(1 + 2h)^{\frac{1}{h}}$, let $t = h$, $a = 2$, so $\lim_{h \to 0}(1 + 2h)^{\frac{1}{h}}=e^{2}$.

Step 2: Analyze the denominator limit

For the denominator $\lim_{h \to 0}2e^{2 + h}$, we can use the limit rule of exponential function. $\lim_{h \to 0}e^{2+h}=e^{2 + 0}=e^{2}$, so $\lim_{h \to 0}2e^{2 + h}=2e^{2}$.

Step 3: Divide the two limits

Now we divide the limit of the numerator by the limit of the denominator: $\frac{\lim_{h \to 0}(1 + 2h)^{\frac{1}{h}}}{\lim_{h \to 0}2e^{2 + h}}=\frac{e^{2}}{2e^{2}}$.

Step 4: Simplify the fraction

Simplify $\frac{e^{2}}{2e^{2}}$, the $e^{2}$ terms cancel out, and we get $\frac{1}{2}$.

Step 1: Rewrite $\cot 3x$

Recall that $\cot\theta=\frac{\cos\theta}{\sin\theta}$, so $\cot 3x=\frac{\cos 3x}{\sin 3x}$. Then the limit becomes $\lim_{x \to \frac{\pi}{2}}\frac{\frac{\cos 3x}{\sin 3x}}{\cos x}=\lim_{x \to \frac{\pi}{2}}\frac{\cos 3x}{\sin 3x\cos x}$.

Step 2: Substitute $x = \frac{\pi}{2}$ into the functions (check for indeterminate form)

First, calculate the values of the functions at $x=\frac{\pi}{2}$:

• $\cos(3\times\frac{\pi}{2})=\cos(\frac{3\pi}{2}) = 0$
• $\sin(3\times\frac{\pi}{2})=\sin(\frac{3\pi}{2})=- 1$
• $\cos(\frac{\pi}{2}) = 0$

So we have a $\frac{0}{0}$ indeterminate form, and we can use L'Hopital's Rule. L'Hopital's Rule states that if $\lim_{x \to a}\frac{f(x)}{g(x)}$ is $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\lim_{x \to a}\frac{f(x)}{g(x)}=\lim_{x \to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}$.
Let $f(x)=\cos 3x$ and $g(x)=\sin 3x\cos x$.

Step 3: Find the derivative of $f(x)$

Using the chain rule, $f^{\prime}(x)=- 3\sin 3x$.

Step 4: Find the derivative of $g(x)$

Using the product rule $(uv)^\prime = u^\prime v+uv^\prime$, where $u = \sin 3x$ and $v=\cos x$.

• $u^\prime = 3\cos 3x$
• $v^\prime=-\sin x$

So $g^{\prime}(x)=3\cos 3x\cos x+\sin 3x(-\sin x)=3\cos 3x\cos x-\sin 3x\sin x$.

Step 5: Apply L'Hopital's Rule

Now we find $\lim_{x \to \frac{\pi}{2}}\frac{f^{\prime}(x)}{g^{\prime}(x)}=\lim_{x \to \frac{\pi}{2}}\frac{-3\sin 3x}{3\cos 3x\cos x-\sin 3x\sin x}$.

Step 6: Substitute $x = \frac{\pi}{2}$ into the new limit

• $\sin(3\times\frac{\pi}{2})=-1$
• $\cos(3\times\frac{\pi}{2}) = 0$
• $\cos(\frac{\pi}{2}) = 0$
• $\sin(\frac{\pi}{2}) = 1$

Substitute these values in:
\[

$$\begin{align*} \frac{-3\times(-1)}{3\times0\times0-(-1)\times1}&=\frac{3}{0 + 1}\\ &=3 \end{align*}$$

\]

Answer:

$\frac{1}{2}$

Problem 37: