Question

Use graphs and tables to find the limit and identify any vertical asymptotes of

Answer 1

@KeithAfasCalcLover

Answer 2

In general for a function of the form:
\[f(x)=\frac{1}{x-d}+c\] Which is just a translated version of \(f(x)=\frac{1}{x}\), the limits exist like so:
\[\lim_{x\to d^-}f(x)=-\infty\textit{ and...}\lim_{x\to d^+}f(x)=\infty\]
So in our case, we have the \(d\) value as 3 so therefore:
\[\lim_{x\to3^-}=-\infty\]

Answer 3

So, -∞ ; x = -3?

Answer 4

Pardon me?

Answer 5

As x approaches 3 from the negative side, f(x) approaches - infinity

Answer 6

Is -∞ ; x = -3 the correct answer

Answer 7

Ehh oh I get it... no it would be:
\[-\infty;x=3\]

Answer 8

Oh, okay. Why 3, and not -3?

Answer 9

In accordance with the same function, if we had the function
\[f(x)=\frac{1}{x-d}+c\]
We know that \(x\neq d\) because then we end up with an undefined value right? So then we know that with the function:
\[f(x)=\frac{1}{x-3};x\neq3\]

Answer 10

Okay.