Limits to Infinity: Exponential Function

In summary: I don't think I'm supposed to use L'Hopital's Rule for these, but I can't think of any other ways to find the limits. Can anyone help?In summary, we can use L'Hopital's Rule to find the limits of e^(x-1)/x, cos(x)-1/sin(x), and other similar functions. However, for functions such as (x+sqrt(x^2+3))/(-2x-s
  • #1
BloodyFrozen
353
1
If we have an exponential fuction,

(for example)

Limx->∞ e(x2+2x+1)/(x2-3)

Would we first determine the limit of the "argument" (not sure if right word) of ex and then replace the "argument" with the limit and then evaluate it?

So for the example above,

The limit of (x2+2x+1)/(x2-3) would be 1 as it approaches ∞.

Then we would evaluate e1.


Would this process be correct?:smile:
 
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  • #2
Hi BloodyFrozen! :smile:

Yes, this is correct. In fact, for every continuous function f, we have

[tex]\lim_{x\rightarrow a}{f(g(x))}=f\left(\lim_{x\rightarrow a}{g(x)}\right)[/tex]

In your example, we have [itex]f(x)=e^x[/itex] and [itex]g(x)=\frac{x^2+2x+1}{x^2-3}[/itex].
 
  • #3
Ok,

And also for limx->-∞ex2/(x-3)

The g(x) part of f(g(x))

equals -∞ at -∞

so,

e-∞ would "theorectically" approach 0 right?
 
  • #4
Yes, that limit is indeed 0! :smile:
 
  • #5
OK, and one last question:-p

limx->-∞ (x2+2x+1)/(-x2+4x+4)

Could we evaluate each one seprately?

top -> -∞
bottom ->∞

so "technically" could we say -∞/∞ = -1

So would the limit be -1?

[P.S. I do know about horizontal asymptotes (Algebra way and Calculus way)]
 
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  • #6
Yes, the limit is -1. It seems you understand these things quite well! :smile:

However, I must say that I don't like

[tex]\frac{\infty}{-\infty}=-1[/tex]

although it's clear what you mean. I wouldn't write this on a test or something. A better way of writing this would be

[tex]\lim_{x\rightarrow -\infty}{\frac{x^2+2x+1}{-x^2+4x+1}}=\lim_{x\rightarrow -\infty}{\frac{x^2}{-x^2}}=\lim_{x\rightarrow -\infty}{-1}=-1[/tex]
 
  • #7
BloodyFrozen said:
limx->-∞ (x2+2x+1)/(-x2+4x+4)

So would the limit be -1?

[P.S. I do know about horizontal asymptotes (algebra way and Calculus way)]

Yes, but not for the reason that you stated. If you're still not convinced, look at:

limx->-∞ (x2)/(-x)

When the highest power on the top and bottom is the same, you simply take the coefficients of the highest power and divide. In your example, it would be 1/-1 = -1
 
  • #8
BloodyFrozen said:
top -> -∞
bottom ->∞

so "technically" could we say -∞/∞ = -1

Not generally. Consider lim (x)/(x^2) as x -> ∞. "Top over bottom" is ∞/∞ but it does not equal 1; it is 0.

To solve lim (x^2+2x+1)/(-x^2+4x+4) as x -> ∞, I'd multiply the argument by [tex]\frac{1/x^2}{1/x^2}[/tex] to get
[tex]\lim_{x \to \infty} \frac{1 + 2/x + 1/x^2}{-1 + 4/x + 4/x^2}[/tex] which is much more clear.
 
  • #9
So what we do would be factor the largest x-exponent from the denominator. That way we would get the other "parts" to m/xn (where m and n are any numbers) as it approaches +/-∞ it'd equal zero. That way it'd be x2/-x2 like you guys said. Am I correct?
 
  • #10
gb7nash said:
Yes, but not for the reason that you stated. If you're still not convinced, look at:

limx->-∞ (x2)/(-x)

When the highest power on the top and bottom is the same, you simply take the coefficients of the highest power and divide. In your example, it would be 1/-1 = -1

Yes, I know.:smile:

I mentioned the lower leveled class (algebra?) way.

If n/m,

If power are same, take coefficients of largest power from numerator and denominator and divide n/m

If n>m, there is no horizontal asymptote.

If n<m, horizontal asymptote at y=0

:smile:
 
  • #11
BloodyFrozen said:
So what we do would be factor the largest x-exponent from the denominator. That way we would get the other "parts" to m/xn (where m and n are any numbers) as it approaches +/-∞ it'd equal zero. That way it'd be x2/-x2 like you guys said. Am I correct?

Yes, that's good!
 
  • #12
BloodyFrozen said:
Yes, I know.:smile:

I mentioned the lower leveled class (algebra?) way.

If n/m,

If power are same, take coefficients of largest power from numerator and denominator and divide n/m

If n>m, there is no horizontal asymptote.

If n<m, horizontal asymptote at y=0

:smile:

The good thing about calculus is that you can now forget all this nonsense :biggrin: Just evaluate things like Unit did. Mathematicians don't like memorizing.
 
  • #13
Ok, Spivak's 4th Edition is doing well so far.:-p

BTW are there any "challenging" limits that I might want to be aware of?
 
  • #14
Oh, you want challenges, hey. :devil:

[tex]\lim_{x\rightarrow +\infty}{\frac{x+\sqrt{x^2+3}}{-2x-\sqrt{2x^2-4}}}[/tex]

[tex]\lim_{x\rightarrow 0}{\frac{e^x-1}{x}}[/tex]

[tex]\lim_{x\rightarrow +\infty}{\frac{\sin(x)}{x}}[/tex]

[tex]\lim_{x\rightarrow 0}{\frac{\cos(x)-1}{\sin(x)}}[/tex]

There are not difficult, but they require some more thought than just plug-and-chug nonsense.
 
  • #15
micromass said:
The good thing about calculus is that you can now forget all this nonsense :biggrin: Just evaluate things like Unit did. Mathematicians don't like memorizing.
Good point!
 
  • #16
Yay, no -b/(2a).
micromass, I'll attempt them tomorrow.
No TeX on phones.

More cont on next page:)
 
  • #17
Also, what is the first and second value in an ordered pair called? I remember they had a specific math "name"
 
  • #18
BloodyFrozen said:
Also, what is the first and second value in an ordered pair called? I remember they had a specific math "name"

The abscissa and the ordinate. Although I've never used those terms in my life.
 
  • #19
Thanks to all.
Goodnight.







Atleast to this thread.;)
 
  • #20
micromass said:
Oh, you want challenges, hey. :devil:

[tex]\lim_{x\rightarrow +\infty}{\frac{x+\sqrt{x^2+3}}{-2x-\sqrt{2x^2-4}}}[/tex]

[tex]\lim_{x\rightarrow 0}{\frac{e^x-1}{x}}[/tex]

[tex]\lim_{x\rightarrow +\infty}{\frac{\sin(x)}{x}}[/tex]

[tex]\lim_{x\rightarrow 0}{\frac{\cos(x)-1}{\sin(x)}}[/tex]

There are not difficult, but they require some more thought than just plug-and-chug nonsense.

I'm having trouble with 1 and 3, but I think I got 2 and 4.

For 2 and 4, I used L'Hopital's Rule

2. (e0-1)/0 = 0/0 ->Differentiate top and bottom
ex = e0 = 1:biggrin:
4. (cox(x)-1)/(sin(x)) = 0/0 Differentiate
-sin(x)/cos(x) = -0/1 = 0:smile:

I can't think of any other ways to find the limits of the above and I'm stuck on numbers 1 and 3:cry:
 
  • #21
Oh, but Hopitals rule is cheating :biggrin:
 
  • #22
micromass said:
Oh, but Hopitals rule is cheating :biggrin:

Yeah, I know:-p

Any hints on how to do 2 and 4 w/o L'hopital's rule?
 
  • #23
BloodyFrozen said:
Yeah, I know:-p

Any hints on how to do 2 and 4 w/o L'hopital's rule?

For (4), you'll have to use the well-known limit

[tex]\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}[/tex]

I actually see no easy way to solve (2) without L'hopital's rule or Taylor series. I just included that one because I'm evil.
 
  • #24
micromass said:
The abscissa and the ordinate. Although I've never used those terms in my life.
Until now!
 
  • #25
micromass said:
For (4), you'll have to use the well-known limit

[tex]\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}[/tex]

I actually see no easy way to solve (2) without L'hopital's rule or Taylor series. I just included that one because I'm evil.

Oh, I remember.
It equals 1, but the problem you gave before was approaching infinity.
The last one reminds me of limx->0 (cos(x)-1)/x = 0, but I'm not sure if that applies here.
 
  • #26
BloodyFrozen said:
Oh, I remember.
It equals 1, but the problem you gave before was approaching infinity.
The last one reminds me of limx->0 (cos(x)-1)/x = 0, but I'm not sure if that applies here.

Yes, the same reasoning applies here. Essentially, you've got a cosine, but you want a sine. You can easily change a cos2 into a sine, but you don't got a square. How do you introduce cos2 into the problem?
 
  • #27
Wait, are we talking about 3 or 4?
 
  • #28
BloodyFrozen said:
Wait, are we talking about 3 or 4?

About 4.

For 3, you'll need to use the squeeze theorem.
 
  • #29
micromass said:
About 4.

For 3, you'll need to use the squeeze theorem.

Ok, first number 3.

-1 =< sin(x) =< 1, therefore

-1/x =< sin(x)/x =< 1/x

Which means the limit is zero. Wow can't believe I forgot that...

EDIT: By any chance would 4 use cofunction identities?
 
  • #30
BloodyFrozen said:
Ok, first number 3.

-1 =< sin(x) =< 1, therefore

-1/x =< sin(x)/x =< 1/x

Which means the limit is zero. Wow can't believe I forgot that...

That's ok! :smile:
 
  • #31
Four is killing me...

Confunction identities? <- nvm don't think that's it
 
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  • #32
multiply numerator and denominator by cos(x)+1.
 
  • #33
Ohhhhhhh, I tried that, but never thought about plugging in 0...fml

sin(x)/(cos(x)+1) ->

0/2=0

....wow (again)
 
  • #34
solution to 2 without using hospital or Taylor:
limx→inf (1+1/x)^x = e
<=> limx→inf xln(1+1/x) =1
<=> limx→0 ln(1+x)/x = 1
<=> limx→0 x - ln(1+x) = 0
<=> limx→0 (ex-1)/x = e^0 =1
 
  • #35
NeroKid said:
solution to 2 without using hospital or Taylor:
limx→inf (1+1/x)^x = e
<=> limx→inf xln(1+1/x) =1
<=> limx→0 ln(1+x)/x = 1
<=> limx→0 x - ln(1+x) = 0
<=> limx→0 (ex-1)/x = e^0 =1

I'm not quite following you. I know that (1+1/n)n = e as n->infinity
 

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