Evaluating a limit with variable in the exponent For

Marzadri9lyy

Marzadri9lyy

Answered question

2022-04-05

Evaluating a limit with variable in the exponent
For limx(12x)x2
I have to use the L'Hospital"s rule, right? So I get:
limxx2log(12x)
And what now? I need to take the derivative of the log, is it: 112x but since there is x, I need to use the chain rule multiply by the derivative of x2?

Answer & Explanation

Charlie Haley

Charlie Haley

Beginner2022-04-06Added 14 answers

Recall the limit:
limy(1+ay)y=ea
I trust you can finish it from here, by an appropriate choice of a and y.
mhapo933its

mhapo933its

Beginner2022-04-07Added 9 answers

If you insist in working with the l'Hospital:
limxx2log(12x)=limxlog(12x)2x=l'Hlimx2x2112x2x2=limx112x=1
Thus, the limit is ,e1,...

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