Practice Final #2

First, notice that as $t \to \infty$, the expression inside the parentheses approaches 1, while the exponent approaches $\infty$. This gives the indeterminate form $1^\infty$, which often suggests that the limit involves the exponential function, $e$. Our goal is to manipulate the expression to fit the standard form for the definition of $e$. We can do this by rewriting the base, $\frac{t+4}{t-1}$, into the form $1 + \text{something}$.


By rewriting $t+4$ as $t-1+5$, we can split the fraction. This isolates a '1' which is crucial for the next step.

$$\begin{align*}$$
& \lim_{t \to \infty} \left(\frac{t+4}{t-1}\right)^{t} \\
&= \lim_{t \to \infty} \left(\frac{t-1+5}{t-1}\right)^{t} \\
&= \lim_{t \to \infty} \left(\frac{t-1}{t-1} + \frac{5}{t-1}\right)^{t} \\
&= \lim_{t \to \infty} \left(1 + \frac{5}{t-1}\right)^{t}
\end{align*}

The key formula to remember is the limit definition of the exponential function: $\lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n = e^x$. Our entire strategy is to make our limit look like this form.
Now that our limit looks very similar to the definition, we can use a substitution to make it match perfectly.


Let $n = t-1$. As $t \to \infty$, it's also true that $n \to \infty$. From this substitution, we also have $t = n+1$. After substituting, we can split the exponent using the property $a^{b+c} = a^b \cdot a^c$. This allows us to separate the expression into two parts. The first part will match the definition of $e^x$ exactly, and the second part will be a simple limit to evaluate.

$$\begin{align*}$$
& \text{Let } n = t-1. \text{ As } t \to \infty, n \to \infty. \\
& \text{Also, } t = n+1. \\
& \lim_{n \to \infty} \left(1 + \frac{5}{n}\right)^{n+1} \\
&= \lim_{n \to \infty} \left[ \left(1 + \frac{5}{n}\right)^{n} \cdot \left(1 + \frac{5}{n}\right)^{1} \right] \\
&= \left[ \lim_{n \to \infty} \left(1 + \frac{5}{n}\right)^{n} \right] \cdot \left[ \lim_{n \to \infty} \left(1 + \frac{5}{n}\right) \right] \\
&= (e^5) \cdot (1+0) \\
&= \boxed{e^5}
\end{align*}

We need to find $f'(x)$ by taking the derivative of each term:

$$\begin{align*}$$
& f(x) = 3\sin(x) - \cos(3x) \\
& f'(x) = 3\cos(x) - (-\sin(3x) \cdot 3) \\
& f'(x) = 3\cos(x) + 3\sin(3x)
\end{align*}



Now we take the derivative of $f'(x)$ to find $f''(x)$:


$$\begin{align*}$$
& f'(x) = 3\cos(x) + 3\sin(3x) \\
& f''(x) = 3(-\sin(x)) + 3(\cos(3x) \cdot 3) \\
& f''(x) = -3\sin(x) + 9\cos(3x)
\end{align*}




When taking derivatives of trigonometric functions with coefficients in the argument, remember the chain rule: $\frac{d}{dx}[\sin(ax)] = a\cos(ax)$ and $\frac{d}{dx}[\cos(ax)] = -a\sin(ax)$.



Now we substitute $x = \frac{\pi}{4}$ into our expression for $f''(x)$:

$$\begin{align*}$$
& f''(\frac{\pi}{4}) \\
&= -3\sin(\frac{\pi}{4}) + 9\cos(3 \cdot \frac{\pi}{4}) \\
&= -3 \cdot \frac{\sqrt{2}}{2} + 9\cos(\frac{3\pi}{4})
\end{align*}

We know that:
- $\sin(\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$
- $\cos(\frac{3\pi}{4}) = -\frac{\sqrt{2}}{2}$

Therefore:

$$\begin{align*}$$
& f''(\frac{\pi}{4}) \\
&= -3 \cdot \frac{\sqrt{2}}{2} + 9 \cdot (-\frac{\sqrt{2}}{2}) \\
&= -\frac{3\sqrt{2}}{2} - \frac{9\sqrt{2}}{2} \\
&= -\frac{12\sqrt{2}}{2} \\
&= -6\sqrt{2}
\end{align*}

Therefore:

$$\boxed{f''(\frac{\pi}{4}) = -6\sqrt{2}}$$

We already found that:

$$\begin{align*}$$
& f'(x) \\
&= \frac{2(x - 1)(x + 2) - (x - 1)^2}{(x + 2)^2} \\
&= \frac{(x - 1)[2(x + 2) - (x - 1)]}{(x + 2)^2} \\
&= \frac{(x - 1)[2x + 4 - x + 1]}{(x + 2)^2} \\
&= \frac{(x - 1)(x + 5)}{(x + 2)^2}
\end{align*}



Setting the derivative equal to zero:

$$\begin{align*}$$
& \frac{(x - 1)(x + 5)}{(x + 2)^2} = 0
\end{align*}

Since the denominator is never zero on our interval, we have:
$(x - 1)(x + 5) = 0$

This gives us $x = 1$ or $x = -5$. Since $-5$ is outside our interval $[0,5]$, we only consider $x = 1$.



We need to check the function value at $x = 0$, $x = 1$, and $x = 5$.

At $x = 0$:

$$\begin{align*}$$
& f(0) \\
&= \frac{(0 - 1)^2}{0 + 2} \\
&= \frac{1}{2} \\
&= 0.5
\end{align*}

At $x = 1$:

$$\begin{align*}$$
& f(1) \\
&= \frac{(1 - 1)^2}{1 + 2} \\
&= \frac{0}{3} \\
&= 0
\end{align*}

At $x = 5$:

$$\begin{align*}$$
& f(5) \\
&= \frac{(5 - 1)^2}{5 + 2} \\
&= \frac{16}{7} \\
&\approx 2.29
\end{align*}



Comparing all three values:
- $f(0) = \frac{1}{2} = 0.5$
- $f(1) = 0$
- $f(5) = \frac{16}{7} \approx 2.29$

The maximum value is $\frac{16}{7}$ which occurs at $x = 5$.

$$\boxed{\text{Maximum value} = \frac{16}{7}}$$

For rational functions, the behavior may be very different on different intervals. Always check all critical points within the interval AND both endpoints!

We already found the derivative of this function in our previous question:

$$\begin{align*}$$
& \frac{dy}{dx} \\
&= \frac{2(x - 1)(x + 2) - (x - 1)^2}{(x + 2)^2}
\end{align*}

To find critical points, we set this equal to zero:

$$\begin{align*}$$
& \frac{2(x - 1)(x + 2) - (x - 1)^2}{(x + 2)^2} = 0
\end{align*}

Since the denominator cannot be zero on our interval (as $x + 2 > 0$ when $x \geq -1.9$), we only need:

$$\begin{align*}$$
& 2(x - 1)(x + 2) - (x - 1)^2 = 0 \\
&(x - 1)[2(x + 2) - (x - 1)] = 0 \\
&(x - 1)[2x + 4 - x + 1] = 0 \\
&(x - 1)[x + 5] = 0
\end{align*}

So $x = 1$ or $x = -5$ are our critical points.



Our interval is $[-8, -4]$, and we found one critical point at $x = -5$ which is within this interval.

Extreme Value Theorem tells us that the maximum must occur either at a critical point within the interval or at an endpoint.

We need to evaluate the function at $x = -8$, $x = -5$, and $x = -4$.



$$\begin{align*}$$
& f(-8) \\
&= \frac{(-8 - 1)^2}{-8 + 2} \\
&= \frac{(-9)^2}{-6} \\
&= \frac{81}{-6} \\
&= -13.5
\end{align*}



$$\begin{align*}$$
& f(-5) \\
&= \frac{(-5 - 1)^2}{-5 + 2} \\
&= \frac{(-6)^2}{-3} \\
&= \frac{36}{-3} \\
&= -12
\end{align*}



$$\begin{align*}$$
& f(-4) \\
&= \frac{(-4 - 1)^2}{-4 + 2} \\
&= \frac{(-5)^2}{-2} \\
&= \frac{25}{-2} \\
&= -12.5
\end{align*}

When finding extreme values on a closed interval, always check both the critical points inside the interval AND the endpoint values!



Comparing all three values:
- $f(-8) = -13.5$
- $f(-5) = -12$
- $f(-4) = -12.5$

The maximum value is $-12$ which occurs at $x = -5$.

$$\boxed{\text{Maximum value} = -12}$$

Let $f(x) = \frac{\tan(x)}{x^2 - 2x + 5}$. Find $f'(0)$.



Since we have a fraction, we'll use the quotient rule.

If $f(x) = \frac{g(x)}{h(x)}$, then:

$$\begin{align*}$$
f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}
\end{align*}



Let's set:
- $g(x) = \tan(x)$
- $h(x) = x^2 - 2x + 5$



$$\begin{align*}$$
& g'(x) = \frac{d}{dx}[\tan(x)] \\
&= \sec^2(x)
\end{align*}

$$\begin{align*}$$
& h'(x) = \frac{d}{dx}[x^2 - 2x + 5] \\
&= 2x - 2
\end{align*}



$$\begin{align*}$$
& f'(x) \\
&= \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2} \\
&= \frac{\sec^2(x)(x^2 - 2x + 5) - \tan(x)(2x - 2)}{(x^2 - 2x + 5)^2}
\end{align*}


When evaluating at x = 0, remember that $\tan(0) = 0$ and $\sec^2(0) = 1$, which simplifies our work considerably!



At $x = 0$:


$$\begin{align*}$$
& f'(0) \\
&= \frac{\sec^2(0)(0^2 - 2(0) + 5) - \tan(0)(2(0) - 2)}{(0^2 - 2(0) + 5)^2} \\
&= \frac{1 \cdot 5 - 0 \cdot (-2)}{5^2} \\
&= \frac{5}{25} \\
&= \frac{1}{5}
\end{align*}


Therefore:

$$\boxed{f'(0) = \frac{1}{5}}$$

Let's check what happens when we try to evaluate the limit directly:

$$\lim_{y \to 0^+} [\cos(2y)]^{1/y^2}$$

As $y \to 0^+$:
- $\cos(2y) \to \cos(0) = 1$
- $1/y^2 \to \infty$

So we have the indeterminate form $1^{\infty}$, which we can't evaluate directly.



To handle this indeterminate form, I'll define:

$$L = \lim_{y \to 0^+} [\cos(2y)]^{1/y^2}$$

Taking the natural logarithm of both sides:

$$\ln(L) = \ln\left(\lim_{y \to 0^+} [\cos(2y)]^{1/y^2}\right)$$

$$\ln(L) = \lim_{y \to 0^+} \ln\left([\cos(2y)]^{1/y^2}\right)$$

Using the property of logarithms:

$$\ln(L) = \lim_{y \to 0^+} \frac{1}{y^2} \ln[\cos(2y)]$$

$$\ln(L) = \lim_{y \to 0^+} \frac{\ln[\cos(2y)]}{y^2}$$



Now we need to evaluate:

$$\lim_{y \to 0^+} \frac{\ln[\cos(2y)]}{y^2}$$

As $y \to 0^+$:
- $\ln[\cos(2y)] \to \ln(1) = 0$
- $y^2 \to 0$

So we have the indeterminate form $\frac{0}{0}$, which means we can apply L'Hôpital's Rule.

Finding the derivatives:

$$\frac{d}{dy}[\ln(\cos(2y))] = \frac{1}{\cos(2y)} \cdot (-\sin(2y) \cdot 2)$$




$$\frac{d}{dy}[\ln(\cos(2y))] = \frac{-2\sin(2y)}{\cos(2y)}$$

$$\frac{d}{dy}[\ln(\cos(2y))] = -2\tan(2y)$$

$$\frac{d}{dy}[y^2] = 2y$$

Applying L'Hôpital's Rule:

$$\lim_{y \to 0^+} \frac{\ln[\cos(2y)]}{y^2} = \lim_{y \to 0^+} \frac{-2\tan(2y)}{2y}$$

$$\lim_{y \to 0^+} \frac{\ln[\cos(2y)]}{y^2} = \lim_{y \to 0^+} \frac{-\tan(2y)}{y}$$



For this new limit:

$$\lim_{y \to 0^+} \frac{-\tan(2y)}{y}$$

As $y \to 0^+$:
- $\tan(2y) \to \tan(0) = 0$
- $y \to 0$

We again have the indeterminate form $\frac{0}{0}$, so we apply L'Hôpital's Rule again.

Finding the derivatives:

$$\frac{d}{dy}[-\tan(2y)] = -\sec^2(2y) \cdot 2$$

$$\frac{d}{dy}[-\tan(2y)] = -2\sec^2(2y)$$

$$\frac{d}{dy}[y] = 1$$

Applying L'Hôpital's Rule:

$$\lim_{y \to 0^+} \frac{-\tan(2y)}{y} = \lim_{y \to 0^+} \frac{-2\sec^2(2y)}{1}$$

$$\lim_{y \to 0^+} \frac{-\tan(2y)}{y} = \lim_{y \to 0^+} -2\sec^2(2y)$$

As $y \to 0^+$:
- $\sec^2(2y) \to \sec^2(0) = 1$

So:

$$\lim_{y \to 0^+} -2\sec^2(2y) = -2 \cdot 1 = -2$$



We found that $\ln(L) = -2$, so:

$$L = e^{-2}$$

For indeterminate forms like $1^{\infty}$, always try the strategy of taking the natural logarithm first, then applying L'Hôpital's Rule to the resulting expression.

Therefore:
$$\boxed{{\displaystyle \underset{y\to 0^{+}}{\lim }[\cos (2y){]}^{1\mathrm{/}{y}^2}=e^{-2}}}$$
\boxed{\lim_{y \to 0^+} [\cos(2y)]^{1/y^2} = e^{-2}}