Practice Final #6

To evaluate this limit, start by simplifying the expression within the tangent function. We’ll try factoring the numerator and the denominator.
The numerator $x^3 - x^2 - x - 2$ can be factored by grouping terms.



To factor $x^3 - x^2 - x - 2$, we group the terms in pairs:

$$x^3-x^2-x-2=(x^3-x^2)-(x+2)$$
x^3 - x^2 - x - 2 = (x^3 - x^2) - (x + 2)


Now, factor out $x^2$ from the first group and factor out $-1$ from the second group:

$$=x^2(x-1)-1(x-1)$$
= x^2(x - 1) - 1(x - 1)


Now we have a common factor of $(x - 2)$:

$$=(x-2)(x^2+x+1)$$
= (x - 2)(x^2 + x + 1)


The denominator $x^2 + 17x - 38$ factors as:

$$x^2+17x-38=(x-2)(x+19)$$
x^2 + 17x - 38 = (x - 2)(x + 19)


So, the expression becomes:

$$\tan \left(\frac{(x-2)(x^2+x+1)}{(x-2)(x+19)}\pi \right)$$
\tan \left( \frac{(x - 2)(x^2 + x + 1)}{(x - 2)(x + 19)} \pi \right)




Now, cancel the common factor $(x - 2)$ from both the numerator and the denominator:

$$\tan \left(\frac{x^2+x+1}{x+19}\pi \right)$$
\tan \left( \frac{x^2 + x + 1}{x + 19} \pi \right)




Now that we’ve simplified the expression, we can substitute $x = 2$ directly:

$$\tan \left(\frac{2^2+2+1}{2+19}\pi \right)=$$
\tan \left( \frac{2^2 + 2 + 1}{2 + 19} \pi \right) =


$$\tan \left(\frac{4+2+1}{21}\pi \right)=$$
\tan \left( \frac{4 + 2 + 1}{21} \pi \right) =


$$\tan \left(\frac{7}{21}\pi \right)=$$
\tan \left( \frac{7}{21} \pi \right) =


$$\tan \left(\frac{\pi }{3}\right)$$
\tan \left( \frac{\pi}{3} \right)


Tip: You need to know the unit circle and the values of common trigonometric functions at special angles, such as $\frac{\pi}{3}$, to solve problems like this.


Since $\tan \left( \frac{\pi}{3} \right) = \sqrt{3}$, we have:

$$\boxed{{\displaystyle \sqrt{3}}}$$
\boxed{\sqrt{3}}




This question is designed to test your ability to handle limits involving trigonometric functions, especially by simplifying rational expressions through factoring. The professor likely included this question to emphasize the importance of **factoring and cancelling terms** before evaluating limits, a common approach in calculus to avoid indeterminate forms.

To differentiate $\frac{1 + e^x}{x^2 + 9}$, we use the quotient rule, which states:

Quotient Rule: If $f(x) = \frac{u}{v}$, then $f'(x) = \frac{u'v - uv'}{v^2}$

In this case:
- $u = 1 + e^x$
- $v = x^2 + 9$



**Derivative of** $u = 1 + e^x$:

$$u^{\mathrm{\prime }}=\frac{d}{dx}(1+e^x)=e^x$$
u' = \frac{d}{dx} (1 + e^x) = e^x


**Derivative of** $v = x^2 + 9$:

$$v^{\mathrm{\prime }}=\frac{d}{dx}(x^2+9)=2x$$
v' = \frac{d}{dx} (x^2 + 9) = 2x




Now, substitute $u$, $u'$, $v$, and $v'$ into the quotient rule formula:







Distribute $e^x$ and $2x$:

$$=\frac{e^x\cdot x^2+9e^x-2x-2x{e}^x}{(x^2+9{)}^2}$$
= \frac{e^x \cdot x^2 + 9e^x - 2x - 2x e^x}{(x^2 + 9)^2}


$$\boxed{{\displaystyle \frac{e^x\cdot x^2+9e^x-2x-2x{e}^x}{(x^2+9{)}^2}}}$$
\boxed{\frac{e^x \cdot x^2 + 9e^x - 2x - 2x e^x}{(x^2 + 9)^2}}

For the natural logarithm $\ln(y)$, the argument $y$ must be greater than zero. In this case that means:

$${\cot }^{-1}(x^2)-\frac{\pi }{6}>0$$
\cot^{-1}(x^2) - \frac{\pi}{6} > 0


which is therefore our first condition that we need to satisfy.



The function $\cot^{-1}(x^2)$ is defined for all $x$ because $x^2 \geq 0$ for all real values of $x$.

So there are no added restrictions on the domain from $\cot^{-1}(x^2)$.





To satisfy the condition for the logarithm, we need:

$${\cot }^{-1}(x^2)-\frac{\pi }{6}>0$$
\cot^{-1}(x^2) - \frac{\pi}{6} > 0


This simplifies to:

$${\cot }^{-1}(x^2)>\frac{\pi }{6}$$
\cot^{-1}(x^2) > \frac{\pi}{6}




Now, we find the values of $x$ such that $\cot^{-1}(x^2) > \frac{\pi}{6}$.

Since $\cot^{-1}(x^2)$ decreases as $x^2$ increases, we need:

$$x^2<\cot \left(\frac{\pi }{6}\right)$$
x^2 < \cot\left( \frac{\pi}{6} \right)


From the unit circle, we know $\cot\left( \frac{\pi}{6} \right) = \sqrt{3}$, so:

$$x^2<\sqrt{3}$$
x^2 < \sqrt{3}


Taking the square root of both sides:

$$\mathrm{\mid }x\mathrm{\mid }<\sqrt[4]{3}$$
|x| < \sqrt[4]{3}


Thus, the domain of the function is:

$$\boxed{{\displaystyle -\sqrt[4]{3}<x<\sqrt[4]{3}}}$$
\boxed{-\sqrt[4]{3} < x < \sqrt[4]{3}}




This problem tests your understanding of the domain of composite functions, especially when involving inverse trigonometric functions and logarithms. The professor likely included this question to ensure you know how to work with domain restrictions and analyze inequalities involving inverse trigonometric expressions.