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February 8th, 2013, 12:04 AM   #1
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Can I apply L'hopital here?

hi all

Can i apply lhopital when the limit of the numerator is different than the denominator but both are infinite?
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February 8th, 2013, 01:24 AM   #2
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Re: Can I apply L'hopital here?

I think so. However, I've been wrong about this before so I'm going to give you an alternative here. My teacher never taught us Lhopitals rule. She took a different approach.
She drew the graphs of the following functions: y=ln(x); y=sqrt(x); y=x^2; y=e^x as an example. Then we looked at the graphs and saw that as x aproaches infinity all these functions approach infinity.
But ln(x) approaches infinity the slowest. If you draw it you will see the function (y=ln(x)) grows very slowly compared to y=sqrt(x). So y=ln(x) approaches infinity a lot slower than y = sqrt(x)
So we ranked these functions based on how fast they grow when x->infinity.
y=ln(x) is the slowest, followed by y=sqrt(x) then y=x, after that goes y=x^2 (or x^n) and the fastest growing function is y=e^x (or a^x where a>1)
But this only applies when x->infinity, not when x->0 or any other number.

we can say 1/x = t, so the the limit becomes the following:

as t-> infinity ln(t) grows slower, and t grows faster therefor we have a smaller infinity divided by a bigger infinity and so we will get 0

We can also try to do it with L'Hopitals rule:

So yes you can do it with L'Hopitals rule
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