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L'Hopital's rule for solving limit

  • Thread starter mousesgr
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  • #1
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1. lim [(1+x)^(1/x) - e ] / x
x ->0

2. lim [sin(2/x)+cos(1/x)]^x
x -> inf

help....
 
Last edited:

Answers and Replies

  • #2
HallsofIvy
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Homework should be posted in the homework forum- and you should show us what you have tried to do. Have you considered L'Hopital's rule?
 
  • #3
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qs 1 is not 0/0 or inf/inf from
how do consider L'Hopital's rule???
 
  • #4
HallsofIvy
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Do you KNOW "L'Hopital's rule"? It is specifically for limit problems where you get things like these.

If you have [tex]lim_{x->a} \frac{f(x)}{g{x}}[/tex] where f(a)= 0 and g(a)= 0, then the limit is the same as [tex]lim_{x->a}\frac{\frac{df}{dx}}{\frac{dg}{dx}}[/tex].

If you get things like [tex]0^0[/tex] or [tex]\infty^{\infty}[/tex] (as your second limit), you can take logarithms to reduct to the first case.
 
  • #5
VietDao29
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mousesgr said:
qs 1 is not 0/0 or inf/inf from
how do consider L'Hopital's rule???
#1 is in form 0 / 0.
Since:
[tex]\lim_{x \rightarrow 0} (1 + x) ^ {\frac{1}{x}} = e[/tex]
So the numerator will tend to 0 as x approaches 0. The denominator also tends to 0. So it's 0 / 0.
You can use L'Hopital's rule to solve for #1.
-------------------
#2 is [tex]1 ^ \infty[/tex]
First, you can try to take logs of both sides.
So let [tex]y = \lim_{x \rightarrow \infty} \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x[/tex]. So:
[tex]\ln y = \ln \left\{ \lim_{x \rightarrow \infty} \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x \right\} = \lim_{x \rightarrow \infty} \ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] ^ x[/tex]
[tex]= \lim_{x \rightarrow \infty} x \ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right] = \lim_{x \rightarrow \infty} \frac{\ln \left[ \sin \left( \frac{2}{x} \right) + \cos \left( \frac{1}{x} \right) \right]}{\frac{1}{x}}[/tex]
This is 0 / 0. So again, you can apply L'Hopital's rule to find the limit.
Viet Dao,
 

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