Chapter 2 Review

Here are some select questions that you may find to be on the more difficult side. I have provided solutions in the collapsed boxes. They are not the exact ones found on the review, but the processes are identical.

1. ${\lim }_{x\to \infty }\frac{10x+5}{3-12x}$

Solution 1

Divide each term in both the numerator and denominator by the variable $x$ raised to the highest term

$$\underset{x\to \infty }{\lim }\frac{\frac{10x}{x}+\frac{5}{x}}{\frac{3}{x}-\frac{12x}{x}}$$

$x$ cancels for terms where it is present while the terms left with $x$ in the denominator tend to zero

$$\underset{x\to \infty }{\lim }\frac{10+0}{0-12}$$

We can simplify to get

$$-\frac{5}{6}$$

2. ${\lim }_{x\to -4^{-}}\frac{4x^2-x+1}{x+4}$

Solution 2

We can start with polynomial long division

$$\begin{array}{rl}& 4x-17+\frac{69}{x+4}\\ x+4& \text{enclose}longdiv4x^2-x+1\\ & -(4x^2+16x)\\ & -17x+1\\ & -(-17x-68)\\ & 69\end{array}$$

This then becomes ${\lim }_{x\to -4^{-}}4x-17+\frac{69}{x+4}$

Evaluating the first two terms separately, we get

$$4(-4)-17+69\underset{x\to -4^{-}}{\lim }\frac{1}{x+4}$$ $$=-33+69\underset{x\to -4^{-}}{\lim }\frac{1}{x+4}$$

We can then evaluate at $x\to -4^{-}$

$$-33+69(\frac{1}{0^{-}})$$ $$=-33+69(-\infty )$$

Constants do not make a big impact at infinity, obtaining the answer

$$-\infty$$

3. ${\lim }_{h\to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}$

Solution 3

We can multiply both the numerator and denominator by the conjugate $\sqrt{x+h}+\sqrt{x}$

$$\underset{h\to 0}{\lim }\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}$$

This multiplies to

$$\underset{h\to 0}{\lim }\frac{x+h-x}{h(\sqrt{x+h}+\sqrt{x})}$$

The two $x$ terms subtract out and $h$ can be cancelled, yielding

$$\underset{h\to 0}{\lim }\frac{1}{\sqrt{x+h}+\sqrt{x}}$$

Using direct substitution of $h=0$, the limit simply becomes

$$\frac{1}{\sqrt{x+0}+\sqrt{x}}$$

Which simplifies to

$$\frac{1}{2\sqrt{x}}$$

4. ${\lim }_{x\to 0}\frac{\sin 5x}{x}$

Solution 4

We can use the known limit ${\lim }_{\alpha \to 0}\frac{\sin \alpha }{\alpha }$ by changing the form of the equation. This can be done through multiplying the numerator and denominator by 5

$$\underset{x\to 0}{\lim }\frac{\sin 5x}{x}\cdot \frac{5}{5}$$

Moving the numerator coefficient outside the limit, we get

$$5\underset{x\to 0}{\lim }\frac{\sin 5x}{5x}$$

Letting $5x=\alpha$, we can use our known limit

$$5\underset{\frac{1}{5}\alpha \to 0}{\lim }\frac{\sin \alpha }{\alpha }$$

Which becomes

$$5\cdot 1=5$$