Limits of Functions practice questions — AP Calculus BC

$\dfrac{1 - \cos x}{x^2} = \dfrac{2\sin^2(x/2)}{x^2} = \dfrac{1}{2} \cdot \!\left(\dfrac{\sin(x/2)}{x/2}\right)^2 \to \dfrac{1}{2}$.

Substituting $x = 2$ gives the indeterminate form $$\frac{2^2 - 5(2) + 6}{2^2 - 4} = \frac{0}{0}.$$ Factor both numerator and denominator and cancel the shared $(x-2)$: $$\frac{x^2 - 5x + 6}{x^2 - 4} = \frac{(x-2)(x-3)}{(x-2)(x+2)} = \frac{x-3}{x+2}.$$ Now substitute $x = 2$ into the simplified form: $$\lim_{x \to 2} \frac{x-3}{x+2} = \frac{2-3}{2+2} = -\frac{1}{4}.$$

A standard fundamental limit: $\displaystyle\lim_{x \to 0} \dfrac{\ln(1 + x)}{x} = 1$.

$\dfrac{2 - 3/x + 1/x^2}{1 + 5/x^2} \to \dfrac{2}{1} = 2$.

A standard fundamental limit: $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x} = 1$.

A standard fundamental limit: $\displaystyle\lim_{x \to 0} \dfrac{e^x - 1}{x} = 1$.

For polynomials, substitute directly: $4 + 6 - 1 = 9$.

By the standard limit $\lim_{t\to 0}\tfrac{\sin t}{t}=1$, we get $\lim_{x\to 0}\tfrac{\sin 3x}{x} = 3\cdot 1 = 3$.

$\dfrac{(x-2)(x+2)}{x-2} = x + 2 \to 4$ as $x \to 2$.

Substitute: $\dfrac{1 + 1}{1 + 2} = \dfrac{2}{3}$.