Practice Questions: Limits

$$\begin{align*} \lim_{x \to 3}\dfrac{x^2-9}{x+3} &=\lim_{x \to 3}\dfrac{(x-3)(x+3)}{x+3}\&=\lim_{x \to 3}(x-3)\&=3-3\&=\text{0} \end{align*}$$

$$\begin{align*} \lim_{x \to 3}\dfrac{x+3}{x^{2}+3x} &=\lim_{x \to 3}\dfrac{x+3}{x(x+3)}\&=\lim_{x \to 3}\dfrac{1}{x}\&= \frac{1}{3} \end{align*}$$

$$\begin{align*} \lim_{x \to 2}\dfrac{3x^2-4x}{3-x} &=\lim_{x \to 2}\dfrac{3(2)^2-4(2)}{3-2}\&=\lim_{x \to 2}\dfrac{12-8}{1}\&= \text{4} \end{align*}$$

$$\begin{align*} \lim_{x \to 4}\dfrac{x^2-x-12}{x-4} &=\lim_{x \to 4}\dfrac{(x-4)(x+3)}{x-4}\&=\lim_{x \to 4}(x+3)\&=4+3\&= \text{7} \end{align*}$$

Note: Even though the function is not defined at $x = 4$, the limit as $x$ tends to $4$ does exist and is equal to $7$.

$$\begin{align*} \lim_{x \to 2} \left( 3x+\dfrac{1}{3x} \right) &=6+\dfrac{1}{6}\&= \frac{37}{6} \end{align*}$$

$$\lim_{x\to 0} \frac{1}{x} = \frac{1}{0} \therefore \text{ does not exist}$$

$$\begin{align*} \lim_{y \to 1} \frac{y-1}{y+1} &= \frac{1-1}{1+1} \\ &=\frac{0}{2}\&=0 \end{align*}$$

$$\lim_{y \to 1} \frac{y+1}{y-1} = \frac{1+1}{1-1} \therefore \text{ does not exist}$$

$$\begin{align*} \lim_{h \to 0} \frac{3h + h^{2}}{h} &= \lim_{h \to 0} \frac{h(3 + h)}{h} \\ &= \lim_{h \to 0} (3 + h) \\ &=3 + 0 \\ &= 3 \end{align*}$$

Note: Even though the function is not defined at $h = 0$, the limit as $h$ tends to $0$ does exist and is equal to $3$.

$$\begin{align*} \lim_{h \to 1} \frac{h^{3} - 1 }{h - 1} &= \lim_{h \to 1} \frac{(h-1)(h^{2} + h + 1) }{h - 1}\&= \lim_{h \to 1} (h^{2} + h + 1) \\ &= 1^{2} + 1 + 1 \\ &= 3 \end{align*}$$

Note: Even though the function is not defined at $h = 1$, the limit as $h$ tends to $1$ does exist and is equal to $3$.

$$\begin{align*} \lim_{x \to 3} \frac{\sqrt{x} - \sqrt{3}}{x - 3} &= \lim_{x \to 3} \frac{\sqrt{x}-\sqrt{3}}{x-3} \times \frac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}}\&= \lim_{x \to 3} \frac{(x-3)}{(x-3)(\sqrt{x}+\sqrt{3})} \\ &= \frac{1}{\sqrt{x}+\sqrt{3}} \\ &= \frac{1}{\sqrt{3}+\sqrt{3}} \\ &= \frac{1}{2\sqrt{3}} \\ &= \frac{\sqrt{3}}{6} \end{align*}$$

Note: The function is not defined at $x = 3$, but the limit as $x$ tends to $3$ does exist and is equal to $\frac{\sqrt{3}}{6}$.