Limits Review

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

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

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

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   $\left(x+3\right)=\ \left(-1+3\right)=2$  .

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   $\frac{1}{x-1}=\frac{1}{2-1}=1$  .

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.

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

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

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