Limits Review

Evaluate the the limit.

 $\lim_{x\longrightarrow-4}\left(x^2+2x-3\right)$  

(a)  

Evaluate the the limit.

 $\lim_{x\longrightarrow5}-\sqrt{x+4}$  

(a)  

Answer explanation

 $\lim_{x\rightarrow5}-\sqrt{x+4}=-\sqrt{5+4}=-\sqrt{9}=-3$  

Evaluate the the limit.

 $\lim_{x\longrightarrow5}\ \ \frac{x+4}{x^2+10x+25}$  

(a)  

Answer explanation

 $\lim_{x\longrightarrow5}\ \ \frac{x+4}{x^2+10x+25}=\ \frac{5+4}{5^2+10\left(5\right)+25}=\frac{9}{100}$  

Evaluate the the limit.

 $\lim_{x\longrightarrow\frac{\pi}{6}}-\sin\left(2x\right)$  

•  $-\frac{\sqrt{3}}{2}$  

•  $\frac{\sqrt{3}}{2}$  

•  $\frac{1}{2}$  

•  $-\frac{1}{2}$  

Answer explanation

 $\lim_{x\longrightarrow\frac{\pi}{6}}-\sin\left(2x\right)=-\sin\left(2\left(\frac{\pi}{6}\right)\right)=-\sin\frac{\pi}{6}=-\frac{\sqrt{3}}{2}$  

Tags

CCSS.HSF.IF.A.2

Evaluate the limit, if it exists.

 $\lim_{x\longrightarrow-1}\ \frac{x^2+4x+3}{x+1}$  

(a)  

Answer explanation

Direct substitution gives  $\frac{0}{0}$  result.  You must factor the numerator to  $\left(x+3\right)\left(x+1\right)$  first and cancel the binomials  $\left(x+1\right)$  .  Then substitute -1 into the remaining binomial   $\left(x+3\right)=\ \left(-1+3\right)=2$  .

Evaluate the limit, if it exists.

 $\lim_{x\longrightarrow2}\ \frac{x-2}{x^2-3x+2}$  

(a)  

Answer explanation

Direct substitution gives  $\frac{0}{0}$  result.  You must factor the denominator to  $\left(x-2\right)\left(x-1\right)$  first and cancel the binomials  $\left(x-2\right)$  .  Then substitute 2 into the remaining binomial   $\frac{1}{x-1}=\frac{1}{2-1}=1$  .

Evaluate the limit, if it exists.

 $\lim_{x\longrightarrow-2}\ \frac{x}{\frac{1}{2+x}-\frac{1}{2}}$  

• 0

• -1

• 1

• 2

Answer explanation

Wow, a whole lot of algebra to needs to be completed to simplify the function to  $-2\left(2+x\right)$  then substitute the -2 into the function to find the limit as x approaches -2 is 0.

Find the vertical asymptote(s) for the function

 $f\left(x\right)=\frac{x+2}{x^2+2x+1}$  

• x = -1

• x = -1 and x = 2

• x = -2

• x = -1/2

Answer explanation

To find the vertical asymptote set the denominator equal 0 and solve for x.  Asymptotes are lines and should be written as x = -1

Tags

CCSS.HSF-IF.C.7D

Describe the behavior of the function to the left of the vertical asymptote

 $f\left(x\right)=\frac{x+2}{x^2+2x+1}$  

•  $\infty$  
  

•  $-\infty$  
  

• 0

• 5

Answer explanation

 $\lim_{x\rightarrow-1^-}\ f\left(x\right)\ =\ \infty$  

Tags

CCSS.HSF-IF.C.7D

Describe the behavior of the function to the right of the vertical asymptote

 $f\left(x\right)=\frac{x+2}{x^2+2x+1}$  

•  $\infty$  
  

•  $-\infty$  
  

• 0

• 5

Answer explanation

 $\lim_{x\rightarrow-1^+}\ f\left(x\right)\ =\ \infty$  

Tags

CCSS.HSF-IF.C.7D