Question

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

Answer 1

how do i do this one @SithsAndGiggles

Answer 2

"... use tables ..."

This method involves evaluating the function at values that get closer and closer to the point of interest.

Clearly, when \(x=6\), you have an asymptote, right? Because \(\dfrac{1}{6-6}=\dfrac{1}{0}\) is undefined.

To see if a limit exists, you would consider the function's values of \(x\) that approach 6 from both sides. From the left (values of \(x\) less than 6), you would check \(f(5.9)\), \(f(5.99)\), \(f(5.999)\), and so on (though 5.999 is a good stopping point).

From the right (values greater than 6), you'd check \(f(6.1)\), \(f(6.01)\), \(f(6.001)\), and so on (again, three decimal place is a good place to stop).

Answer 3

so like this?

\[f(5.9)=\frac{ 1 }{ 5.9-6 }=\frac{ 1 }{ -.1 }=-10\]
??

Answer 4

Yes, you keep repeating the process. You'll find that with 5.99 and 5.999 and so on, the value gets more and more negative:
\[f(5.99)=-100\\
f(5.999)=-1000\]
which tells you that the function approaches \(-\infty\) from the left.

Answer 5

wait is it supposed to be a -6-6?

Answer 6

No, I'm not sure what you mean.

I've known some teachers/professors to say that infinite limits are nonexistent, so what you say as your answer depends on who's grading.

Answer 7

isn't that a negative six under the lim?

Answer 8

-∞ ; x = 6

-∞ ; x = -6

∞ ; x = -6

1 ; no vertical asymptotes

these are my answers, that's why I asked if it is a -6 or just a 6

Answer 9

No that says \(x\) is approaching 6 from the left.

You found that the limit is \(-\infty\).

The second part is to say what the asymptote is. When is the denominator zero? When \(x=6\).

Answer 10

ooooooooooh ok thank you!!! :)

Answer 11

yw