Question

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Use graphs and tables to find the limit and identify any vertical asymptotes of limit of 1 divided by the quantity x minus 3 as x approaches 3 from the left

∞; x = -3
-∞; x = -3
-∞; x = 3
1 ; no vertical asymptotes

Is it A?

Answer 1

the function is $f(x)= \frac{1}{x-3}$

we are looking at $\lim_{x \to \ 3^{-} } \frac{1}{x-3}$, that is, we are looking at what happens with the function when x are values very very close to 3, from the left.

method 1: using tables

 $\lim_{x \to \ 3^{-} } \frac{1}{x-3}$ is approximately the value of f at a number very very close to 3, but a little less than 3.

this number can be thought as a=2.9999999998

a-3 is negative,since a is very close to 3, but smaller.

$\frac{1}{x-3}$ for x=a is  $\frac{1}{0.0000000002} =20000000000$

so for values of  x closer to 3, $\frac{1}{x-3}$ is larger and larger, in the negative direction.

This means that $\lim_{x \to \ 3^{-} } \frac{1}{x-3}$=-∞ and the asymptote is the vertical line x=3, since for x=3, the function is not defined.

method 2: by graphing the function using a graphic calculator, we see that the graph gets very very close to the vertical line x=3, but never touches it, so this line is a vertical asymptote.

Also, we see that the closer x gets to 3, the smaller the value of f becomes, so  

$\lim_{x \to \ 3^{-} } \frac{1}{x-3}$=-∞ 


Answer:

-∞; x = 3