Compute Limits: f(x) = 2,3 - Does Not Exist

[buffgilville]

Compute the limit of:
1) f(x) = 2(square root of (x^2 - 1))/(x+1)

as x approaches -1 from the right.

2) f(x) = 3(x+2)/(absolute value of x + 2)

as x approaches -2 from the left.

I got does not exist for both because and x>-1 and x<-2, respectively. Am I right?

 

[Leong]

$$\begin{multline*}<br /> \begin{split}<br /> &f(x)=\frac{3(x+2)}{|x+2|}\\<br /> &Approches\ -2\ from\ the\ left:\\<br /> &x+2<0;\ |x+2|=-(x+2)\\<br /> &f(x)=\frac{3(x+2)}{-(x+2)}\\<br /> &f(x) approaches -3.<br /> \end{split}<br /> \end{multline*}$$

 

[M98Ranger]

Yes, you are correct. Both limits do not exist because the function is undefined at the given values of x. As x approaches -1 from the right, the denominator of the first function becomes 0, making the entire fraction undefined. Similarly, as x approaches -2 from the left, the absolute value of x + 2 becomes 0, also making the entire fraction undefined. Therefore, the limits do not exist in both cases.