find any vertical asymptotes of the function limit of 1 divided by the quantity x minus 8 as x approaches 8 from the left

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find any vertical asymptotes of the function limit of 1 divided by the quantity x minus 8 as x approaches 8 from the left

Mathematics
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https://i.gyazo.com/600783c4843f253829c56f7ccc1a8a91.png
\[f(x)=\frac{1}{x-8}\]?
yea.

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Other answers:

you know what it looks like?
not exactly
has a vertical asymptote at \(x=8\) for sure
did you graph these in some pre-calc class?
this is for my final assignment for my online class
actually we can cheat easily, do it without knowing the graph
oh well if it is on line lets cheat like crazy
http://www.wolframalpha.com/input/?i=1%2F%28x-8%29
you and i would become good friends lol
you can see that as \(x\to 8^-\) it goes to \(-\infty\)
you can see that from the picture right?
yea
ok my motto "cheating is learning"
awesome. that is now my new motto
all rights reserved
wait so is x=8 the only one?
yes of course
only place where the denominator would be zero
ok thank you so much
yw
and is it ok if you can help me with one last question
ok one last question
Find the fifth roots of 243(cos 240° + i sin 240°)
lord i thought it as going to be a short one
ok first the fifth root of 243 is 3 i think
lol sorry.
then divide \(240\) by \(5\)
i get 48
that would be 48
yea
so one answer is \[ 3(\cos (48°) + i \sin (48°))\]
next we add 360 to 240 and divide by 5 again
ok so it would be 120
\[360+240=600,600\div 5=120\] right
so 2nd answer is \[3(\cos (120°) + i \sin (120°))\]
lather, rinse, repeat
alright. so once again i add 360 and divide by 5?
\[600+360=960, 960\div 5=192\]
3(cos(192°)+isin(192°))
third answer is \[3(\cos (192°) + i \sin (192°))\] you are too fast for me
haha sorry
no that is good, you get the picture, even though evidently you will never use this again
i will let you finish the last two \[960+360\] etc
ok thanks so much for the help. thats everything i needed help with
you are welcome, hope you pass

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