Trigonometric Limit Practice Problems

If $-x^4\le f(x)\le x^4$ near zero, find $\lim_{x\to0}f(x)$.

If $2-x^2\le g(x)\le2+x^2$ near zero, find the limit.

Evaluate $\lim_{x\to0}x^3\sin(1/x)$.

Evaluate $\lim_{x\to0}x^2\cos(7/x)$.

Evaluate $\lim_{x\to0}|x|\sin(4/x)$.

Evaluate $\lim_{x\to0}x\cos(1/x^2)$.

Suppose $|f(x)|\le5|x-2|$. Find $\lim_{x\to2}f(x)$.

Explain why boundedness of $f$ alone is not enough to conclude $\lim_{x\to0}f(x)=0$.

$\displaystyle\lim_{x\to0}\frac{\sin x}{x}$

$\displaystyle\lim_{x\to0}\frac{\sin(2x)}{x}$

$\displaystyle\lim_{x\to0}\frac{\sin(7x)}{3x}$

$\displaystyle\lim_{x\to0}\frac{\sin(5x)}{\sin(2x)}$

$\displaystyle\lim_{x\to0}\frac{\sin(3x)}{\sin(8x)}$

$\displaystyle\lim_{t\to0}\frac{\sin(\pi t)}{t}$

$\displaystyle\lim_{x\to0}\frac{x}{\sin(4x)}$

$\displaystyle\lim_{x\to0}\frac{3x}{\sin(5x)}$

$\displaystyle\lim_{x\to0}\frac{\tan x}{x}$

$\displaystyle\lim_{x\to0}\frac{\tan(4x)}{x}$

$\displaystyle\lim_{x\to0}\frac{\tan(3x)}{2x}$

$\displaystyle\lim_{x\to0}\frac{1-\cos x}{x}$

$\displaystyle\lim_{x\to0}\frac{1-\cos x}{x^2}$

$\displaystyle\lim_{x\to0}\frac{1-\cos(4x)}{x^2}$

$\displaystyle\lim_{x\to0}\frac{\sin^2x}{x^2}$

$\displaystyle\lim_{x\to0}\frac{\sin^2(3x)}{x^2}$

$\displaystyle\lim_{x\to0}\frac{\sin(2x)\sin(3x)}{x^2}$

$\displaystyle\lim_{x\to0}\frac{\tan(5x)}{\sin(2x)}$

$\displaystyle\lim_{x\to0}\frac{1-\cos(2x)}{\sin^2x}$

$\displaystyle\lim_{x\to0}\frac{\sin(4x)}{x\cos x}$

$\displaystyle\lim_{x\to0}\frac{\sin x}{x(1+\cos x)}$

$\displaystyle\lim_{x\to0}\frac{\tan x-\sin x}{x}$

$\displaystyle\lim_{x\to0}\frac{1-\cos x}{x\sin x}$

$\displaystyle\lim_{x\to0}\frac{\sin(3x)}{\tan(5x)}$

Explain why radians are required for $\lim_{x\to0}\sin x/x=1$.

Derive $\lim_{x\to0}\tan x/x=1$ from the sine limit.

Derive $\lim_{x\to0}(1-\cos x)/x^2=1/2$.

A student writes $\sin(5x)=5x$. Explain what is wrong and how to use the limit correctly.

Give a squeeze argument for $\lim_{x\to0}x^4\sin(1/x^3)$.

Create a trigonometric limit near zero whose value is $7/4$.