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Limit as x approaches Infinity

  • Thread starter caesius
  • Start date
  • #1
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Homework Statement


Evaluate [tex]lim (1+a/x)^x[/tex]
(that's limit as x tends to +infinity, sorry don't know how to latex that)

Homework Equations





The Attempt at a Solution


Stumped. Having that [tex]x[/tex] exponent has my confused. As x tends to infinity I know what's inside the brackets will tend to one, but the exponent will make it tend to infinity. But the question has three marks to it, it can't be that simple.
 

Answers and Replies

  • #2
lurflurf
Homework Helper
2,423
123
so
call your limit
f(a)
-show the the limit exists (the a=1 case should be familar)
-show f(x)f(y)=f(x+y)
-show f is continous and differentiable at some point
now you should be able to identify f
 
  • #3
Caesius, have you seen all definitions of e?

Oh and your latex that you wanted to do is \lim_{x \rightarrow \infty} which becomes [tex]\lim_{x \rightarrow \infty}[/tex]
 
  • #4
1,752
1
[tex]\lim_{x\rightarrow\infty}\left(1+\frac a x\right)^x[/tex]

Set it equal to y, take the natural log of both sides.

[tex]y=\left(1+\frac a x\right)^x[/tex]

[tex]\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}x\ln\left(1+\frac a x\right)[/tex]

[tex]\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}x\ln\left(1+\frac a x\right)[/tex]

Now you have an Indeterminate form of [tex]\infty\cdot0[/tex]

Keep solving till you can apply L'Hopital's Rule on the right side, then you will have "solved for [tex]\ln y[/tex], so use algebra to solve for "y" and you're pretty much done.
 
  • #5
lurflurf
Homework Helper
2,423
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Caesius, have you seen all definitions of e?
QUOTE]

No one has seen all definitions of e as there are an infinite number of them.
 
  • #6
[tex]\lim_{x\rightarrow\infty}\left(1+\frac a x\right)^x[/tex]

Set it equal to y, take the natural log of both sides.

[tex]y=\left(1+\frac a x\right)^x[/tex]

[tex]\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}x\ln\left(1+\frac a x\right)[/tex]

[tex]\lim_{x\rightarrow\infty}\ln y=\lim_{x\rightarrow\infty}x\ln\left(1+\frac a x\right)[/tex]

Now you have an Indeterminate form of [tex]\infty\cdot0[/tex]

Keep solving till you can apply L'Hopital's Rule on the right side, then you will have "solved for [tex]\ln y[/tex], so use algebra to solve for "y" and you're pretty much done.

Thats a great solution but I think no need to go that long;

as x approaches infinity "(a/x)" equates 0 .

Hence : 1+0 = 1
 
  • #7
Gib Z
Homework Helper
3,346
4
Thats a great solution but I think no need to go that long;

as x approaches infinity "(a/x)" equates 0 .

Hence : 1+0 = 1
And that is just wrong. Try reading the other posts, they did mention "e" several times for a good reason.
 
  • #8
1,752
1
Thats a great solution but I think no need to go that long;

as x approaches infinity "(a/x)" equates 0 .

Hence : 1+0 = 1
Indeterminate Power of form: [tex]1^{\infty}[/tex] so yes it is necessary.
 

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