High School
SAT
ACT
University
MCAT
LSAT
DAT
Wize University Calculus 1 Textbook > Limits

Limit Laws

BasicsPrevious Section
One-Sided LimitsNext Section
Popular Courses
Find My Course
0:00 / 0:00

Limit Laws

We can compute limits more easily by using the following limit laws.

If lim⁡x→af(x)=L,lim⁡x→ag(x)=M \displaystyle\lim_{x\rightarrow a}f(x)=L, \displaystyle\lim_{x\rightarrow a}g(x)=M , and kk is a constant, then:
  1. lim⁡x→a[f(x)+g(x)]=L+M\displaystyle\lim_{x\to a}\left[f\left(x\right)+ g\left(x\right)\right]=L+M
  2. lim⁡x→a[f(x)−g(x)]=L−M\displaystyle\lim_{x\to a}\left[f\left(x\right)- g\left(x\right)\right]=L-M
  3. lim⁡x→a[f(x)g(x)]=LM\displaystyle\lim_{x\to a}[f\left(x\right)g\left(x\right)]=L M
  4. lim⁡x→a kf(x)=kL\displaystyle\lim_{x\to a}\ kf\left(x\right)=kL
  5. lim⁡x→a[f(x)g(x)]=LM,  (M≠0)\displaystyle\lim_{x\to a} \left[\frac{f\left(x\right)}{g\left(x\right)}\right] = \frac{L}{M},\ \ (M\neq0)
  6. lim⁡x→a[f(x)]mn=Lmn,  (L>0 if n is even, and L≠0 if m/n<0)\displaystyle\lim_{x\to a}\left[f\left(x\right)\right]^{\frac{m}{n}}=L^{\frac{m}{n}},\ \ (L>0\ \text{if }n\ \text{is even, and }L\neq0\ \text{if }m/n<0)
  7. If f(x)≤g(x)f(x)\le g(x) for every xx in an interval containing aa, then L≤ML\le M

Wize Tip
Direct Substitution
Always first try to substitute the value that xx approaches directly into the expression → if you get a number, that's the value of the limit!

Evaluate the limit:

lim⁡x→3[xx2−5+1]2\displaystyle\lim_{x\rightarrow3}\left[\frac{x}{x^2-5}+1\right]^2

Extra Practice
Limit Laws
lim⁡x→π2 cos⁡x−πsin⁡x=\displaystyle \lim_{x\to\frac{\pi}{2}}\ \frac{\cos x-\pi}{\sin x}=
Limit with Absolute Value
lim⁡x→0 ∣3−2x∣−∣2x−3∣x=\displaystyle\lim_{x\to0}\ \frac{\left|3-2x\right|-\left|2x-3\right|}{x}=
Limit Laws
If lim⁡x→∞cos⁡(arctan⁡(x2−2))f(x)=π\displaystyle \lim_{x\to\infty}\frac{\cos\left(\arctan\left(x^2-2\right)\right)}{f\left(x\right)}=\pi, then lim⁡x→∞cos⁡(f(x))\displaystyle \lim_{x\to\infty}\cos\left(f\left(x\right)\right) is
Limit Laws
lim⁡x→π2 cos⁡x−πsin⁡x=\displaystyle \lim_{x\to\frac{\pi}{2}}\ \frac{\cos x-\pi}{\sin x}=
Basic limit techniques - Absolute value
Evaluate lim⁡x→4−∣x−4∣x2−4.\lim\limits_{x\rightarrow4^-}\frac{\vert x-4\vert}{x^2-4}.
Limit Laws
lim⁡x→π2 cos⁡x−πsin⁡x=\displaystyle \lim_{x\to\frac{\pi}{2}}\ \frac{\cos x-\pi}{\sin x}=
Basic limit techniques - Absolute value
Evaluate lim⁡x→4−∣x−4∣x2−4.\lim\limits_{x\rightarrow4^-}\frac{\vert x-4\vert}{x^2-4}.
Evaluating limits
Determine the limit
lim⁡ x→0 ∣5x−2∣−∣5x+2∣2x\lim\ _{x\rightarrow0\ }\frac{\left|5x-2\right|-\left|5x+2\right|}{2x}
Limit Law & Direct Substitution
If lim⁡x→π f(x)sin⁡2x=−13\displaystyle \lim_{x\to\pi}\ \frac{f\left(x\right)}{\sin^2x}=-\frac{1}{3}, then what is the value of lim⁡x→π f(x)sin⁡x\displaystyle \lim_{x\to\pi}\ \frac{f\left(x\right)}{\sin x}?
Limit Laws
If lim⁡x→∞cos⁡(arctan⁡(x2−2))f(x)=π\displaystyle \lim_{x\to\infty}\frac{\cos\left(\arctan\left(x^2-2\right)\right)}{f\left(x\right)}=\pi, then lim⁡x→∞cos⁡(f(x))\displaystyle \lim_{x\to\infty}\cos\left(f\left(x\right)\right) is
Limit Laws
lim⁡x→π2 cos⁡x−πsin⁡x=\displaystyle \lim_{x\to\frac{\pi}{2}}\ \frac{\cos x-\pi}{\sin x}=
Basic limit techniques - Absolute value
Evaluate lim⁡x→4−∣x−4∣x2−4.\lim\limits_{x\rightarrow4^-}\frac{\vert x-4\vert}{x^2-4}.
Determine the value of lim⁡x→0(4+x)2−16x\displaystyle \lim_{x\rightarrow 0}\frac{(4+x)^2-16}{x}.
Determine the value of lim⁡x→(π/2)+sin⁡2(x)−xsin⁡x\displaystyle \lim_{x\rightarrow(\pi/2)^+}\frac{\sin^2 (x) -x}{\sin x}.
Evaluate the limit, lim⁡x→0(sin⁡x+x+2)(cos⁡x+1)\lim_{x\rightarrow 0}(\sin x+x+2)(\cos x+1).
Evaluate the limit, lim⁡x→2(x2+2x+1)\lim_{x\rightarrow 2}(x^2+2x+1).
Limit Law & Direct Substitution
If lim⁡x→π f(x)sin⁡2x=−13\displaystyle \lim_{x\to\pi}\ \frac{f\left(x\right)}{\sin^2x}=-\frac{1}{3}, then what is the value of lim⁡x→π f(x)sin⁡x\displaystyle \lim_{x\to\pi}\ \frac{f\left(x\right)}{\sin x}?
Find the value of b such that lim⁡x→0ax+b−5x=2\lim_{x\rightarrow0}\frac{\sqrt{ax+b-5}}{x}=2.
Limits: Indeterminate Forms
If lim⁡x →2 4−2 f(x)2−x\displaystyle\lim_{x\ \rightarrow2}\ \frac{4-2\ f\left(x\right)}{2-x} is defined, what is the value of lim⁡x →2  f(x)\displaystyle\lim_{x\ \rightarrow2}\ \ f\left(x\right) ?
Evaluating limits
Determine the limit
lim⁡ x→0 ∣5x−2∣−∣5x+2∣2x\lim\ _{x\rightarrow0\ }\frac{\left|5x-2\right|-\left|5x+2\right|}{2x}
Limit with Absolute Value
lim⁡x→0 ∣3−2x∣−∣2x−3∣x=\displaystyle\lim_{x\to0}\ \frac{\left|3-2x\right|-\left|2x-3\right|}{x}=
Limits: Basics
Evaluate the limit:
lim⁡x→3[xx2−5+1]2\displaystyle\lim_{x\rightarrow3}\left[\frac{x}{x^2-5}+1\right]^2