Limits to Infinity: Exponential Function

[BloodyFrozen]

If we have an exponential fuction,

(for example)

Lim_x->∞ e^{(x2+2x+1)/(x2-3)}

Would we first determine the limit of the "argument" (not sure if right word) of e^x 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?

 

[micromass]

Hi BloodyFrozen!

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

\lim_{x\rightarrow a}{f(g(x))}=f\left(\lim_{x\rightarrow a}{g(x)}\right)

In your example, we have f(x)=e^x and g(x)=\frac{x^2+2x+1}{x^2-3}.

 

[BloodyFrozen]

Ok,

And also for lim_x->-∞e^x2/(x-3)

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

equals -∞ at -∞

so,

e^-∞ would "theorectically" approach 0 right?

 

[micromass]

Yes, that limit is indeed 0!

 

[BloodyFrozen]

OK, and one last question

lim_x->-∞ (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)]

 

[micromass]

Yes, the limit is -1. It seems you understand these things quite well!

However, I must say that I don't like

\frac{\infty}{-\infty}=-1

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

\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

 

[gb7nash]

lim_x->-∞ (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:

lim_x->-∞ (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

 

[Unit]

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 \frac{1/x^2}{1/x^2} to get
\lim_{x \to \infty} \frac{1 + 2/x + 1/x^2}{-1 + 4/x + 4/x^2} which is much more clear.

 

[BloodyFrozen]

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?

 

[BloodyFrozen]

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

lim_x->-∞ (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.

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

 

[micromass]

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!

 

[micromass]

Yes, I know.

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

The good thing about calculus is that you can now forget all this nonsense Just evaluate things like Unit did. Mathematicians don't like memorizing.

 

[BloodyFrozen]

Ok, Spivak's 4^th Edition is doing well so far.

BTW are there any "challenging" limits that I might want to be aware of?

 

[micromass]

Oh, you want challenges, hey.

\lim_{x\rightarrow +\infty}{\frac{x+\sqrt{x^2+3}}{-2x-\sqrt{2x^2-4}}}

\lim_{x\rightarrow 0}{\frac{e^x-1}{x}}

\lim_{x\rightarrow +\infty}{\frac{\sin(x)}{x}}

\lim_{x\rightarrow 0}{\frac{\cos(x)-1}{\sin(x)}}

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

 

[Mark44]

The good thing about calculus is that you can now forget all this nonsense Just evaluate things like Unit did. Mathematicians don't like memorizing.

Good point!

 

[BloodyFrozen]

Yay, no -b/(2a).
micromass, I'll attempt them tomorrow.
No TeX on phones.

More cont on next page:)

 

[BloodyFrozen]

Also, what is the first and second value in an ordered pair called? I remember they had a specific math "name"

 

[micromass]

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.

 

[BloodyFrozen]

Thanks to all.
Goodnight.







Atleast to this thread.;)

 

[BloodyFrozen]

Oh, you want challenges, hey.

\lim_{x\rightarrow +\infty}{\frac{x+\sqrt{x^2+3}}{-2x-\sqrt{2x^2-4}}}

\lim_{x\rightarrow 0}{\frac{e^x-1}{x}}

\lim_{x\rightarrow +\infty}{\frac{\sin(x)}{x}}

\lim_{x\rightarrow 0}{\frac{\cos(x)-1}{\sin(x)}}

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
e^x = e⁰ = 1
4. (cox(x)-1)/(sin(x)) = 0/0 Differentiate
-sin(x)/cos(x) = -0/1 = 0

I can't think of any other ways to find the limits of the above and I'm stuck on numbers 1 and 3

 

[micromass]

Oh, but Hopitals rule is cheating

 

[BloodyFrozen]

Oh, but Hopitals rule is cheating

Yeah, I know

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

 

[micromass]

Yeah, I know

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

\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}

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.

 

[Mark44]

The abscissa and the ordinate. Although I've never used those terms in my life.

Until now!

 

[BloodyFrozen]

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

\lim_{x\rightarrow 0}{\frac{\sin(x)}{x}}

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 lim_x->0 (cos(x)-1)/x = 0, but I'm not sure if that applies here.

 

[micromass]

Oh, I remember.
It equals 1, but the problem you gave before was approaching infinity.
The last one reminds me of lim_x->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?

 

[BloodyFrozen]

Wait, are we talking about 3 or 4?

 

[micromass]

Wait, are we talking about 3 or 4?

About 4.

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

 

[BloodyFrozen]

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?

 

[micromass]

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!

 

[BloodyFrozen]

Four is killing me...

Confunction identities? <- nvm don't think that's it

 

[micromass]

multiply numerator and denominator by cos(x)+1.

 

[BloodyFrozen]

Ohhhhhhh, I tried that, but never thought about plugging in 0...fml

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

0/2=0

....wow (again)

 

[NeroKid]

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

 

[BloodyFrozen]

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

 

[NeroKid]

u take the ln at the both side of limx --> inf (1+1/x)^x = e
and from there u switch to x' which equal to 1/x

 

[BloodyFrozen]

How did you get from here:
<=> limx→inf xln(1+1/x) =1
to here?:
<=> limx→0 ln(1+x)/x = 1

 

[NeroKid]

limx→inf xln(1+1/x) =1
<=> lim(1/x)→0 xln(1+1/x)=1
call x' = 1/x
limx'→0 ln(1+x')/x' =1

 

[BloodyFrozen]

limx→inf xln(1+1/x) =1
<=> lim(1/x)→0 xln(1+1/x)=1
call x' = 1/x
limx'→0 ln(1+x')/x' =1

Ok, I see.
Thanks

 

[BloodyFrozen]

Lastly,

How did you get these two steps?

<=> limx→0 ln(1+x')/x' = 1
<=> limx→0 x - ln(1+x) = 0

 

[NeroKid]

because limx→0 ln(1+x)/x = 1 and limx→0 ln(1+x) = 0 and limx→0 x =0
so limx→0 ln(1+x) = limx→0 x

 

[BloodyFrozen]

Got it. Thanks