Question

PLEASE HELP WILL GIVE MEDAL
Use graphs and tables to find the limit and identify any vertical asymptotes of

Answer 1

@SolomonZelman

Answer 2

\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~7}\frac{1}{(x-7)^2}}\)


\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~7}\frac{1^2}{(x-7)^2}}\)


\(\large\color{slate}{\displaystyle\lim_{x \rightarrow ~7}\left( \frac{1}{x-7} \right)^2}\)


\(\large\color{slate}{\displaystyle\left( \lim_{x \rightarrow ~7}~\frac{1}{x-7} \right)^2}\)

Answer 3

I have just applied basic algebra and limit rules.

Answer 4

@SolomonZelman please help me

Answer 5

@SolomonZelman why did u skip me/ ignore me?

Answer 6

Now, do you think that \(\large\color{slate}{\displaystyle \lim_{x \rightarrow ~7}~\frac{1}{x-7} }\) exists or not?

Answer 7

I am not good at math.

Answer 8

@SolomonZelman please please i need help you're the best at math thats why you've helped all those people

Answer 9

I am bad at math. highjumpcollect

Answer 10

the limit does not exist then?

Answer 11

Yes

Answer 12

will u try i know you're the best

Answer 13

Because,

\(\large\color{slate}{\displaystyle \lim_{x \rightarrow ~7^-}~\frac{1}{x-7} =~-~\infty}\)


\(\large\color{slate}{-\infty}\) because as \(\large\color{slate}{x\rightarrow~7^-}\), the denominator will be a very small negative decimal.


\(\large\color{slate}{\displaystyle \lim_{x \rightarrow ~7^+}~\frac{1}{x-7} =~+~\infty}\)


\(\large\color{slate}{-\infty}\) because as \(\large\color{slate}{x\rightarrow~7^+}\), the denominator will be a very small positive decimal.

Answer 14

A case where, \(\large\color{slate}{\displaystyle \lim_{x \rightarrow ~7^-}~\frac{1}{x-7} \ne\lim_{x \rightarrow ~7^+}~\frac{1}{x-7} }\)

Answer 15

This is why there is no limit.

Answer 16

Questions?

Answer 17

Oh for second reasoning there should be a positive, not a negative infinity, excuse me for the typo.