AI TutorHigh School
SAT
ACT
University
MCAT
LSAT
DAT

Basic limit techniques - Absolute value

Related Topics

Wize University Calculus 1 Textbook > Limits

Basics

2 Activities

Wize University Calculus 1 Textbook > Limits

Limit Laws

2 Activities

Evaluate lim⁡x→4−∣x−4∣x2−4.\lim\limits_{x\rightarrow4^-}\frac{\vert x-4\vert}{x^2-4}.
More Basics Questions:
Infinite Limits
Evaluate lim⁡x→∞arctan⁡(x−x3)\displaystyle \lim_{x\to\infty}\arctan\left(x-x^3\right)
Limits: Basics
lim⁡x→0∣2x−1∣−∣2x+1∣x\displaystyle\lim_{x\rightarrow0}\dfrac{\lvert 2x-1\rvert-\lvert 2x+1\rvert}{x}
Infinite Limits
Evaluate lim⁡x→∞arctan⁡(x−x3)\displaystyle \lim_{x\to\infty}\arctan\left(x-x^3\right)
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
lim⁡x→0∣2x−1∣−∣2x+1∣x\displaystyle\lim_{x\rightarrow0}\dfrac{\lvert 2x-1\rvert-\lvert 2x+1\rvert}{x}
Limits: Basics
Determine lim⁡x→π2ln⁡(sin⁡x)\displaystyle \lim_{x\to\frac{\pi}{2}}\ln\left(\sin x\right).
Inverse Trig & Exponential Limits
Find the value of lim⁡x→0+arctan⁡(e1x)\displaystyle\lim_{x\rightarrow0^+}\arctan\left(e^{\frac{1}{x}}\right).
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}.
Inverse Trig & Exponential Limits
Find the value of lim⁡x→0+arctan⁡(e1x)\displaystyle\lim_{x\rightarrow0^+}\arctan\left(e^{\frac{1}{x}}\right).
Inverse Trig & Exponential Limits
Find the value of lim⁡x→0+arctan⁡(e1x)\displaystyle\lim_{x\rightarrow0^+}\arctan\left(e^{\frac{1}{x}}\right).
Limits: Absolute Values
Q.\textbf{Q.} Evaluate lim⁡x→0∣2x−1∣−∣2x+1∣x\displaystyle \lim_{x\rightarrow 0}\frac{\lvert 2x-1\rvert-\lvert 2x+1\rvert}{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}=
Infinite Limits
Evaluate lim⁡x→∞arctan⁡(x−x3)\displaystyle \lim_{x\to\infty}\arctan\left(x-x^3\right)
Inverse Trig & Exponential Limits
Find the value of lim⁡x→0+arctan⁡(e1x)\displaystyle\lim_{x\rightarrow0^+}\arctan\left(e^{\frac{1}{x}}\right).
Trigonometric Functions
Evaluate lim⁡x→0sin⁡(1∣x∣)\lim_{x\rightarrow0}\sin\left(\frac{1}{\left|x\right|}\right).
Limits: Absolute Values
Q.\textbf{Q.} Evaluate lim⁡x→0∣2x−1∣−∣2x+1∣x\displaystyle \lim_{x\rightarrow 0}\frac{\lvert 2x-1\rvert-\lvert 2x+1\rvert}{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}.
Limits: Basics
lim⁡x→0∣2x−1∣−∣2x+1∣x\displaystyle\lim_{x\rightarrow0}\dfrac{\lvert 2x-1\rvert-\lvert 2x+1\rvert}{x}
Limits: Basics
Determine lim⁡x→π2ln⁡(sin⁡x)\displaystyle \lim_{x\to\frac{\pi}{2}}\ln\left(\sin x\right).
lim⁡x→0∣2x−1∣−∣2x+1∣x\displaystyle\lim_{x\rightarrow0}\dfrac{\lvert 2x-1\rvert-\lvert 2x+1\rvert}{x}
More Limit Laws Questions:
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}=
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