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cassieforlife5
 one year ago
Find limit as x approaches 1
cassieforlife5
 one year ago
Find limit as x approaches 1

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cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0dw:1444091846364:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0my guess is that for a first step, get rid of that compound fraction

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0okay could you help me with that?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.0im betting they want \(x\to 0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would make more sense

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0no it's as \[\lim_{x \rightarrow 1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0then without doing a ton of work, you can pretty much forget about a limit existing the denominator of \(\frac{1}{\sqrt{x+1}}\) goes to zero and the numerator does not, so you are not going to have a limit forget i mentioned the algebra, although that is usually a good first step

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0okay I thought the question was really weird so I put nonexistent as the limit before and just wanted to see if I had done something wrong. Thanks so much!!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0btw i guess it is worth mentioning that the domain of this beast is \((1,\infty)\) so you cannot take the limit in any case a limit is two sided and you cannot approach \(1\) from the left

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok that is wrong, because the domain excludes zero as well

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the above is still true

cassieforlife5
 one year ago
Best ResponseYou've already chosen the best response.0yeah I graphed it and it stopped at 1 and only continued onto the right which is why I assumed it was nonexistent
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