Limits of Functions practice questions — AP Calculus BC

${\displaystyle \frac{1-\cos x}{x^2}}={\displaystyle \frac{2{\sin }^2(x\mathrm{/}2)}{x^2}}={\displaystyle \frac{1}{2}}\cdot \text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{\sin (x\mathrm{/}2)}{x\mathrm{/}2}}\right)}^2\to {\displaystyle \frac{1}{2}}$.

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

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

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

A standard fundamental limit: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\sin x}{x}}=1}$.

A standard fundamental limit: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x-1}{x}}=1}$.

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

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

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

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