Evaluating Limits

$\dfrac{9 - 3 - 6}{3 - 3} = \dfrac{0}{0}$, so keep going.

$x^2 - x - 6 = (x - 3)(x + 2)$.

$\displaystyle\lim_{x \to 3} \dfrac{(x - 3)(x + 2)}{x - 3} = \lim_{x \to 3} (x + 2)$.

$3 + 2 = 5$.

$\dfrac{4 - 10 + 6}{4 - 4} = \dfrac{0}{0}$.

Top: $(x + 2)(x + 3)$. Bottom: $(x + 2)(x - 2)$.

$\displaystyle\lim_{x \to -2} \dfrac{x + 3}{x - 2}$.

$\dfrac{-2 + 3}{-2 - 2} = -\dfrac{1}{4}$.

$\dfrac{8 - 8}{4 - 2 - 2} = \dfrac{0}{0}$.

Top: $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$. Bottom: $(x - 2)(x + 1)$.

$\displaystyle\lim_{x \to 2} \dfrac{x^2 + 2x + 4}{x + 1}$.

$\dfrac{4 + 4 + 4}{3} = 4$.

$\dfrac{1 - 2 + 5}{-1 + 3} = \dfrac{4}{2} = 2$. A number, so we are done.

Not needed — plugging in did not give $\dfrac{0}{0}$.

Nothing to cancel.

The answer is the value from step 1.

$\dfrac{\sqrt{9} - 3}{0} = \dfrac{0}{0}$.

$\displaystyle\lim_{x \to 5} \dfrac{\sqrt{x + 4} - 3}{x - 5} \cdot \dfrac{\sqrt{x + 4} + 3}{\sqrt{x + 4} + 3}$.

Top: $(x + 4) - 9 = x - 5$, so we have $\dfrac{x - 5}{(x - 5)(\sqrt{x + 4} + 3)}$.

$\displaystyle\lim_{x \to 5} \dfrac{1}{\sqrt{x + 4} + 3}$.

$\dfrac{1}{3 + 3} = \dfrac{1}{6}$.

$\dfrac{0}{\sqrt{9} - 3} = \dfrac{0}{0}$.

Multiply by $\dfrac{\sqrt{2x + 5} + 3}{\sqrt{2x + 5} + 3}$.

Bottom: $(2x + 5) - 9 = 2x - 4 = 2(x - 2)$, so we have $\dfrac{(x - 2)(\sqrt{2x + 5} + 3)}{2(x - 2)}$.

$\displaystyle\lim_{x \to 2} \dfrac{\sqrt{2x + 5} + 3}{2}$.

$\dfrac{3 + 3}{2} = 3$.

$\dfrac{2 - 2}{0} = \dfrac{0}{0}$.

Multiply by $\dfrac{\sqrt{4 + x} + \sqrt{4 - x}}{\sqrt{4 + x} + \sqrt{4 - x}}$.

Top: $(4 + x) - (4 - x) = 2x$, so we have $\dfrac{2x}{x(\sqrt{4 + x} + \sqrt{4 - x})}$.

$\displaystyle\lim_{x \to 0} \dfrac{2}{\sqrt{4 + x} + \sqrt{4 - x}}$.

$\dfrac{2}{2 + 2} = \dfrac{1}{2}$.

$\dfrac{\frac{1}{3} - \frac{1}{3}}{0} = \dfrac{0}{0}$.

$\dfrac{1}{x} - \dfrac{1}{3} = \dfrac{3 - x}{3x}$.

$3 - x = -(x - 3)$, so the numerator is $\dfrac{-(x - 3)}{3x}$.

$\displaystyle\lim_{x \to 3} \dfrac{-(x - 3)}{3x(x - 3)} = \lim_{x \to 3} \dfrac{-1}{3x}$.

$\dfrac{-1}{9}$.

$\dfrac{\frac{3}{3} - 1}{0} = \dfrac{0}{0}$.

$\dfrac{3}{x + 4} - 1 = \dfrac{3 - (x + 4)}{x + 4} = \dfrac{-1 - x}{x + 4}$.

$-1 - x = -(x + 1)$, so the numerator is $\dfrac{-(x + 1)}{x + 4}$.

$\displaystyle\lim_{x \to -1} \dfrac{-(x + 1)}{(x + 4)(x + 1)} = \lim_{x \to -1} \dfrac{-1}{x + 4}$.

$\dfrac{-1}{3}$.

$\dfrac{\frac{1}{4} - \frac{1}{4}}{0} = \dfrac{0}{0}$.

$\dfrac{1}{(2 + h)^2} - \dfrac{1}{4} = \dfrac{4 - (2 + h)^2}{4(2 + h)^2}$.

$4 - (4 + 4h + h^2) = -4h - h^2 = -h(4 + h)$. Here the bad factor is $h$.

$\displaystyle\lim_{h \to 0} \dfrac{-h(4 + h)}{4(2 + h)^2\, h} = \lim_{h \to 0} \dfrac{-(4 + h)}{4(2 + h)^2}$.

$\dfrac{-4}{4 \cdot 4} = -\dfrac{1}{4}$.

Left of $2$: $x^2 - 1$. Right of $2$: $2x - 1$. The value $f(2) = 0$ is not used.

$\displaystyle\lim_{x \to 2^-} (x^2 - 1) = 4 - 1 = 3$.

$\displaystyle\lim_{x \to 2^+} (2x - 1) = 4 - 1 = 3$.

$3 = 3$, so the limit exists and equals $3$, even though $f(2) = 0$.

$x > 3$: $|x - 3| = x - 3$, so $f(x) = \dfrac{(x - 3)(x + 3)}{x - 3} = x + 3$. $x < 3$: $|x - 3| = -(x - 3)$, so $f(x) = -(x + 3)$.

$\displaystyle\lim_{x \to 3^-} -(x + 3) = -6$.

$\displaystyle\lim_{x \to 3^+} (x + 3) = 6$.

$-6 \neq 6$, so the two-sided limit does not exist.

Left of $1$: the straight line. Right of $1$: the curve. The filled dot is $f(1)$ only.

Following the line toward $x = 1$: height $3$.

Following the curve toward $x = 1$: height $1$.

$3 \neq 1$, so $\displaystyle\lim_{x \to 1} f(x)$ does not exist. Separately, $f(1) = 4$.

Left of $3$: $cx + 1$. Right of $3$: $cx^2 - 8$.

$\displaystyle\lim_{x \to 3^-} (cx + 1) = 3c + 1$.

$\displaystyle\lim_{x \to 3^+} (cx^2 - 8) = 9c - 8$.

Need $3c + 1 = 9c - 8 \Rightarrow 6c = 9 \Rightarrow c = \dfrac{3}{2}$. Then the limit is $3 \cdot \dfrac{3}{2} + 1 = \dfrac{11}{2}$.

$\dfrac{\sin 0}{e^{0} - 1} = \dfrac{0}{0}$.

Top: $u = 5x$. Bottom: $u = 2x$.

$\sin(5x) \approx 5x$ and $e^{2x} - 1 \approx 2x$.

$\dfrac{5x}{2x} = \dfrac{5}{2}$.

$\dfrac{1 - 1}{0} = \dfrac{0}{0}$.

$u = 4x$ inside the cosine.

$1 - \cos(4x) \approx \dfrac{(4x)^2}{2} = 8x^2$.

$\dfrac{8x^2}{x^2} = 8$.

$\dfrac{1 - 1}{\log 1} = \dfrac{0}{0}$.

Top: $u = 8x$. Bottom: $u = -2x$.

$\sqrt{1 + 8x} - 1 \approx \dfrac{8x}{2} = 4x$ and $\log(1 - 2x) \approx -2x$.

$\dfrac{4x}{-2x} = -2$.

$\dfrac{0 \cdot 0}{1 - 1} = \dfrac{0}{0}$.

Top: $u = 3x$ in the sine. Bottom: $u = 2x$ in the cosine.

$x \sin(3x) \approx x \cdot 3x = 3x^2$ and $1 - \cos(2x) \approx \dfrac{(2x)^2}{2} = 2x^2$.

$\dfrac{3x^2}{2x^2} = \dfrac{3}{2}$.

$-1 \le \cos\!\left(\dfrac{2}{x}\right) \le 1$.

$x^4 \ge 0$, so $-x^4 \le x^4 \cos\!\left(\dfrac{2}{x}\right) \le x^4$.

$\displaystyle\lim_{x \to 0} (-x^4) = 0$ and $\displaystyle\lim_{x \to 0} x^4 = 0$.

Both bounds go to $0$, so by the squeeze theorem the limit is $0$.

$-1 \le \sin\dfrac{1}{x} \le 1$, so $2 \le 3 + \sin\dfrac{1}{x} \le 4$.

$x^2 \ge 0$, so $2x^2 \le x^2\left(3 + \sin\dfrac{1}{x}\right) \le 4x^2$.

$2x^2 \to 0$ and $4x^2 \to 0$.

By the squeeze theorem the limit is $0$.

$-1 \le \sin\dfrac{1}{x} \le 1$, and $e^{t}$ is increasing, so $e^{-1} \le e^{\sin(1/x)} \le e$.

$\sqrt{x} \ge 0$, so $\dfrac{\sqrt{x}}{e} \le \sqrt{x}\, e^{\sin(1/x)} \le e\sqrt{x}$.

$\dfrac{\sqrt{x}}{e} \to 0$ and $e\sqrt{x} \to 0$ as $x \to 0^+$.

By the squeeze theorem the limit is $0$.