Finite Limit Practice Problems

$\displaystyle\lim_{x\to3}(2x+7)$

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

$\displaystyle\lim_{t\to4}(t^3-2t)$

$\displaystyle\lim_{x\to1}\frac{x^2+3}{x+2}$

$\displaystyle\lim_{x\to-1}\frac{2x^2+x+5}{x^2+4}$

$\displaystyle\lim_{u\to8}\sqrt[3]{u+19}$

$\displaystyle\lim_{x\to0}(3\cos x+2\sin x)$

If $f(x)\to2$ and $g(x)\to-3$, find $\lim[4f(x)-g(x)]$.

With the same limits, find $\lim f(x)g(x)$.

With the same limits, find $\lim\dfrac{f(x)^2+g(x)}{f(x)-g(x)}$.

State why the quotient law cannot be used when the denominator's limit is zero.

Give an example where direct substitution produces a real number and therefore no algebraic simplification is needed.

$\displaystyle\lim_{x\to5}\frac{x^2-25}{x-5}$

$\displaystyle\lim_{x\to-3}\frac{x^2+5x+6}{x+3}$

$\displaystyle\lim_{x\to4}\frac{x^2-7x+12}{x-4}$

$\displaystyle\lim_{x\to2}\frac{x^3-8}{x-2}$

$\displaystyle\lim_{x\to-2}\frac{x^3+8}{x+2}$

$\displaystyle\lim_{x\to1}\frac{x^4-1}{x-1}$

$\displaystyle\lim_{x\to-1}\frac{x^4-1}{x+1}$

$\displaystyle\lim_{x\to3}\frac{x^2-9}{x^2-4x+3}$

$\displaystyle\lim_{x\to1}\frac{x^3-1}{x^2-1}$

$\displaystyle\lim_{h\to0}\frac{(2+h)^2-4}{h}$

$\displaystyle\lim_{h\to0}\frac{(3+h)^3-27}{h}$

Explain why the cancelled expression and original expression may differ at the target but have the same limit.

$\displaystyle\lim_{x\to0}\frac{\sqrt{x+9}-3}{x}$

$\displaystyle\lim_{x\to0}\frac{\sqrt{x+16}-4}{x}$

$\displaystyle\lim_{x\to25}\frac{x-25}{\sqrt{x}-5}$

$\displaystyle\lim_{x\to4}\frac{\sqrt{x}-2}{x-4}$

$\displaystyle\lim_{x\to0}\frac{\sqrt{1+5x}-1}{x}$

$\displaystyle\lim_{x\to0}\frac{\sqrt{4+x}-\sqrt{4-x}}{x}$

$\displaystyle\lim_{x\to3}\frac{x-3}{\sqrt{x+6}-3}$

$\displaystyle\lim_{x\to0}\frac{\sqrt{1+2x}-\sqrt{1-x}}{x}$

$\displaystyle\lim_{x\to0}\frac{\frac1{x+1}-1}{x}$

$\displaystyle\lim_{x\to0}\frac{\frac1{x+2}-\frac12}{x}$

$\displaystyle\lim_{x\to0}\frac{\frac1{3+x}-\frac13}{x}$

$\displaystyle\lim_{x\to2}\frac{\frac1x-\frac12}{x-2}$

$\displaystyle\lim_{x\to1}\frac{\frac1{x+1}-\frac12}{x-1}$

$\displaystyle\lim_{h\to0}\frac{\frac1{a+h}-\frac1a}{h}$, where $a\ne0$.

$\displaystyle\lim_{x\to0}\frac{\frac{x}{x+1}}{x}$

$\displaystyle\lim_{x\to0}\frac{\frac1{1-x}-1}{x}$

$\displaystyle\lim_{x\to0^-}\frac{|x|}{x}$

$\displaystyle\lim_{x\to0^+}\frac{|x|}{x}$

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

$\displaystyle\lim_{x\to2}\frac{|x-2|}{x-2}$

$\displaystyle\lim_{x\to-1^-}\frac{|x+1|}{x+1}$

$\displaystyle\lim_{x\to3}\frac{|x-3|}{\sqrt{|x-3|}}$

Let $f(x)=\begin{cases}x+2,&x<1,\\x^2+1,&x\ge1.\end{cases}$ Find the limit at $1$.

Let $g(x)=\begin{cases}2x,&x<2,\\x+3,&x\ge2.\end{cases}$ Find both one-sided limits at $2$.

$\displaystyle\lim_{x\to2}\frac{x^3-4x}{x^2-4}$

$\displaystyle\lim_{x\to1}\frac{x^5-1}{x^2-1}$

$\displaystyle\lim_{x\to0}\frac{\sqrt{9+2x}-3}{x}$

$\displaystyle\lim_{x\to4}\frac{\frac1x-\frac14}{x-4}$

$\displaystyle\lim_{x\to0}\frac{|x|}{\sqrt{|x|}}$

A student cancels $x$ in $(x^2+x)/x$ and obtains $x+1$. Explain why this cancellation is valid for $x\ne0$, and distinguish it from cancelling a term across addition without factoring.

Design a $0/0$ limit whose value is $5$.

Design a $0/0$ limit whose value is $0$.

Design a $0/0$ limit that does not exist.

Write a short decision procedure for determining whether to factor, rationalize, combine fractions, or split into one-sided cases.