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Practice: Limit of a Function

Related Topics

Wize University Calculus 1 Textbook > Limits

One-Sided Limits

3 Activities

Example: Limit of a Function

The following is the graph of the function f(x)f\left(x\right)
Determine the following limits:
a. lim⁡x→0f(x)=\displaystyle \lim_{x\to0}f\left(x\right)=

b. i)lim⁡x→2−f(x)=\displaystyle \lim_{x\to2^-}f\left(x\right)=
ii) lim⁡x→2+f(x)=\displaystyle \lim_{x\to2^+}f\left(x\right)=
iii) lim⁡x→2f(x)=\displaystyle \lim_{x\to2}f\left(x\right)=

c. i) lim⁡x→5−f(x)=\displaystyle \lim_{x\to5^-}f\left(x\right)=
ii) lim⁡x→5+f(x)=\displaystyle \lim_{x\to5^+}f\left(x\right)=
iii) lim⁡x→5f(x)=\displaystyle \lim_{x\to5}f\left(x\right)=

d. i) lim⁡x→8−f(x)=\displaystyle \lim_{x\to8^-}f\left(x\right)=
ii) lim⁡x→8+f(x)=\displaystyle \lim_{x\to8^+}f\left(x\right)=
iii) lim⁡x→8f(x)=\displaystyle \lim_{x\to8}f\left(x\right)=

More One-Sided Limits Questions:
One-sided Limits
Evaluate the limit lim⁡x→1x2−1∣x−1∣\displaystyle\lim_{x\to1}\frac{x^2-1}{\left|x-1\right|}
Practice: One-Sided Limits
Find:
a)lim⁡x→2+∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
b) lim⁡x→2−∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
Practice: One-Sided Limits
Find:
a)lim⁡x→2+∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
b) lim⁡x→2−∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
One-sided Limits
Evaluate the limit lim⁡x→1x2−1∣x−1∣\displaystyle\lim_{x\to1}\frac{x^2-1}{\left|x-1\right|}
Limit with Absolute Value
Evaluate lim⁡x→2 x−2∣5x−10∣\displaystyle\lim_{x\to2}\ \frac{x-2}{|5x-10|}, if it exists.
Limit with Absolute Value
Evaluate lim⁡x→2 x−2∣5x−10∣\displaystyle\lim_{x\to2}\ \frac{x-2}{|5x-10|}, if it exists.
Limit with Absolute Value
Evaluate lim⁡x→2 x−2∣5x−10∣\displaystyle\lim_{x\to2}\ \frac{x-2}{|5x-10|}, if it exists.
One-sided Limits: Direct Substitution
Practice: Direct Substitution
Given the following function, evaluate the limits as x approaches -1, 2, and 3.
f(x)={−3x+5if  x<−1−10if  x=−19−x2if  −1<x≤21−∣2x−3∣if  x>2f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
Practice: One-Sided Limits
Find:
a)lim⁡x→2+∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
b) lim⁡x→2−∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
One-sided limits: Evaluating limits by factoring
Practice: Evaluating limits
Evaluate lim⁡x→2−x2−3x+2∣x−2∣\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{\left|x-2\right|}
One-sided Limits: Direct Substitution
Practice: Direct Substitution
Given the following function, what should the limits be as x approaches -1, 2, and 3?
f(x)={−3x+5if  x<−1−10if  x=−19−x2if  −1<x≤21−∣2x−3∣if  x>2f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
Practice: Direct Substitution
Practice: Direct Substitution
Given the following function, evaluate the limits below.
f(x)={−3x+5if  x<−1−10if  x=−19−x2if  −1<x≤21−∣2x−3∣if  x>2f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
Indeterminate Forms: One-sided limits
Evaluate:
lim⁡x→1−∣x−1∣x−1\lim_{x \rightarrow 1^-} \frac{|x-1|}{x-1}
One-Sided Limits
Evaluate
lim⁡x→3−∣x−3∣x2−3\lim\limits_{x\to 3^-} \frac{|x-3|}{x^2-3}
Practice: Direct Substitution
Practice: Direct Substitution
Given the following function, evaluate the limits below.
f(x)={−3x+5if  x<−1−10if  x=−19−x2if  −1<x≤21−∣2x−3∣if  x>2f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
Consider the function f(x)={12x+aifx<28x2+a−1ifx≥2f\left(x\right)=\begin{cases}\dfrac{1}{2}x+a &\text{if} &x<2\\ \dfrac{8}{x^2+a-1}&\text{if} &x\geq2\\ \end{cases} . What are the values of a for which the limit lim⁡x→2f(x)\displaystyle\lim_{x\to2}f\left(x\right) exists?
Limits: Practice
Find the following limit:
lim⁡x→2−5(x+1)x−2\displaystyle\lim_{x\rightarrow 2}\frac{-5(x+1)}{x-2}
Limits: Practice
Find the following limit:
lim⁡x→2−5(x+1)x−2\displaystyle\lim_{x\rightarrow 2}\frac{-5(x+1)}{x-2}
Evaluate the following limits and the value of the function, where the graphs of f(x) and g(x) are given in the figure. If the limit does not exist, input "DNE".
a) lim⁡x→1f(x)=\displaystyle\lim_{x\rightarrow1}f(x)= b)lim⁡x→2−g(x)=\displaystyle\lim_{x\rightarrow2^-}g(x)=
c)lim⁡x→1+f(x)=\displaystyle\lim_{x\rightarrow1^+}f(x)= d)lim⁡x→2+g(x)=\displaystyle\lim_{x\rightarrow2^+}g(x)=
Let us call all the functions in the figure below as y(x)y\left(x\right). Evaluate the following limits, if they exist. If the limit does not exist, write "DNE".
A) lim⁡x→−1 y(x)\displaystyle\lim_{x\rightarrow-1}\ y(x)
B) lim⁡x→2 y(x)\displaystyle\lim_{x\rightarrow2}\ y(x)
Practice: One-Sided Limits
Find:
a)lim⁡x→2+∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
b) lim⁡x→2−∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}
One-sided Limits
Evaluate the limit lim⁡x→1x2−1∣x−1∣\displaystyle\lim_{x\to1}\frac{x^2-1}{\left|x-1\right|}
One-sided Limits: Direct Substitution
Practice: Direct Substitution
Given the following function, evaluate the limits below.
f(x)={−3x+5if  x<−1−10if  x=−19−x2if  −1<x≤21−∣2x−3∣if  x>2f\left(x\right)= \begin{cases} -3x+5 & \text{if } \ x<-1 \\ -10 & \text{if }\ x=-1\\ 9-x^2 & \text{if }\ -1< x\le2\\ 1-|2x-3| & \text{if } \ x>2 \end{cases}
One-sided limits: Evaluating limits by factoring
Practice: Evaluating limits
Evaluate lim⁡x→2−x2−3x+2∣x−2∣\lim_{x\rightarrow2^-}\frac{x^2-3x+2}{\left|x-2\right|}
Limit with Absolute Value
Evaluate lim⁡x→2 x−2∣5x−10∣\displaystyle\lim_{x\to2}\ \frac{x-2}{|5x-10|}, if it exists.
One-Sided Limits
If lim⁡x→a−f(x)=2L+3\lim_{x \rightarrow a^-}f(x) = 2L + 3 and lim⁡x→a+f(x)=3L−1\lim_{x \rightarrow a^+}f(x) = 3L - 1 and lim⁡x→af(x)\lim_{x \rightarrow a}f(x) exists with LL a real number, what is the value of LL ?
Practice: One-Sided Limits
Find:
a)lim⁡x→2+∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^+}\frac{|x-2|}{x-2}
b) lim⁡x→2−∣x−2∣x−2\displaystyle\lim_{x\rightarrow 2^-}\frac{|x-2|}{x-2}