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anonymous
 one year ago
Lim x goes to 1 x^21/x1
anyone can find this limit?
anonymous
 one year ago
Lim x goes to 1 x^21/x1 anyone can find this limit?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x\to 1}\frac{x^21}{x1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0seems unlikely since this is is a piecewise function it is one thing if \(x>1\) and quite another if \(x<1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if \(x>1\) then \(x1=x1\) when you factor and cancel you get \[x+1\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you get something else if \(x<1\) try it and see

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I wonder if I should use left and right limit to do this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you have no choice but to do that, since \(x1\) is a piecewise function which changes definition at \(x=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0lets take it step by step

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if \(x>1\) then \(x1=x1\) right?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0did i lose you there?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0if \(x>1\) then \(x1=x1\)so your function is \[\frac{x^21}{x1}=\frac{(x+1)(x1)}{x1}=x+1\] making the limit as \(x\to 1^+=1+1=2\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I see, now the left limit I guess

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0left limit is different because if \(x<1\) then \(x1=1x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so the limit doesnt exist right

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no the two sided limit does not exist
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