Big brains LOOK UP this small limit problem

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Big brains.... LOOK UP this small limit problm

hi,

Lt (exp)x/x I couldnt solve this in any way;neither my friends could.
x->0


So it is a need for me to ask help. Helllp...
 

Galileo

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Do you mean:

[tex]\lim_{x\to 0}\frac{e^x}{x}[/tex]
?

If you just look where the numerator and denominator tend to, that'll give you enough info.
 

James R

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Another way to look at it is like this:

[tex]e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \frac{x^5}{5!} + ...[/tex]

So

[tex]\frac{e^x}{x} = \frac{1}{x} + 1 + \frac{x}{2!} + \frac{x^2}{3!} + \frac{x^3}{4!} + \frac{x^4}{5!} + ...[/tex]

What is the limit of this as x goes to zero?

All of the terms go to zero except the 1 and the 1/x. What happens to the 1/x?
 
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"All of the terms go to zero except the 1 and the 1/x. What happens to the 1/x?"

That was my problem.Solving in any ways gave me the answer infinity. What to do now?Any other ideas to solve this very simple problm???


BACK-WORD:I am on a bet with my friends that i will do it...Help me else I will lose money :cry:
 
111
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It doesn't have a limit as x goes to zero. One sided limits exist of course. Anuyway you could think about this equation as if it were a k/x equation. Of course This is not the same but, the graphs are similar. Like, as x goes to zero e^x goes to one, and x goes to zero. What happens then?.

Like I told you there's no limit as x goes to zero. You could find the right hand side or the left hand side limit but the two are different. So no limit at 0. The best way to see this graph. As soon as you graph this, everything will be cristal clear.
 
176
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hello there

well you could make a substitution make
[tex]x=\frac{1}{u}}[/tex]
[tex]\lim_{x\to 0}\frac{e^x}{x}=\lim_{u\to\infty}ue^{\frac{1}{u}}=\infty[/tex]
now that should be pretty obvious since
[tex]\lim_{u\to\infty}e^{\frac{1}{u}}=1[/tex]
good luck with the bet

Steven
 

HallsofIvy

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lwrthy said:
"All of the terms go to zero except the 1 and the 1/x. What happens to the 1/x?"

That was my problem.Solving in any ways gave me the answer infinity. What to do now?Any other ideas to solve this very simple problm???
So what's wrong with infinity? This limit does not exist. The function ex/x diverges to infinity as x goes to 0.
 

cronxeh

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but what about L'Hopital?

[tex]\lim_{x\to 0}\frac{e^x}{x}=\frac{(e^x)'}{(x)'}=\frac{e^x}{1}= 1[/tex]

Nah probably cant be right, should be infinity though :frown:


Aaaahh I see my mistake there.

[tex]e^0[/tex] != [tex]0[/tex]
 
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hello there

the e^x does no satisfy the conditions to use L' hospital

steven
 
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I still cannot understand, a function cannot have two limits at a point, and this one has. As you approach to zero from left f(x) goes to minus infinity, from the other side, it goes to infinity. In the first quadrant as soon as you are past 1, the graph is just like k/x.

Thus your answer's wrong steven... Or am I doing something wrong?
 
176
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well you make a substitution make
[tex]x=\frac{1}{u}}[/tex]
[tex]\lim_{x\to 0^+}\frac{e^x}{x}=\lim_{u\to\infty}ue^{\frac{1}{u}}=\infty[/tex]

hello there

well a function can have 2 limits at a point either approaching a limit from the left or the right, such sanarios arise when you have a discontinuity at a point as for the above from my 2nd last post, I showed how the limit approaches it from the right hand side now i will show how the limit approaches from the left hand side

well you make a substitution make
[tex]x=\frac{1}{u}}[/tex]
[tex]\lim_{x\to 0^-}\frac{e^x}{x}=\lim_{u\to-\infty}ue^{\frac{1}{u}}=-\infty[/tex]

steven
 

HallsofIvy

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wisredz said:
I still cannot understand, a function cannot have two limits at a point, and this one has.
If a function has a limit then it must be unique. The point is that [tex]\lim_{x\rightarrow0}\frac{e^x}{x}[/tex] does not exist!
 
I think your friends set you up dude
 
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Just graph it, I bet it will be obvious where it approaches.

As x goes to 0, [itex] e^x [/itex] goes to 1, and the denominator goes to 0, so

[tex] \lim_{x\rightarrow 0^+} \frac{e^x}{x} = \infty [/tex].

I dont see where the problem is.
 
wisredz said:
I still cannot understand, a function cannot have two limits at a point, and this one has. As you approach to zero from left f(x) goes to minus infinity, from the other side, it goes to infinity. In the first quadrant as soon as you are past 1, the graph is just like k/x.

Thus your answer's wrong steven... Or am I doing something wrong?
No, you are right. There is no limit. the limit as x->0 of e^x/x is undefined.
 
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hello there

[tex]\lim_{x\to 0^+}\frac{e^x}{x}\not=\lim_{x\to 0^-}\frac{e^x}{x}[/tex]
hence
[tex]\lim_{x\to 0}\frac{e^x}{x}[/tex]
does not exist since left and right limits do not give a unique limit

steven
 
2,208
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steven187 said:
hello there

[tex]\lim_{x\to 0^+}\frac{e^x}{x}\not=\lim_{x\to 0^-}\frac{e^x}{x}[/tex]
hence
[tex]\lim_{x\to 0}\frac{e^x}{x}[/tex]
does not exist since left and right limits do not give a unique limit

steven
I agree, I thought he was looking at only a right-handed limit.
 
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steven187 said:
hello there

[tex]\lim_{x\to 0^+}\frac{e^x}{x}\not=\lim_{x\to 0^-}\frac{e^x}{x}[/tex]
hence
[tex]\lim_{x\to 0}\frac{e^x}{x}[/tex]
does not exist since left and right limits do not give a unique limit

steven

thanks everybody.
so i should understand that the limit does not exist.is it?
 
176
0
hello there
yes you are right the limit does not exist at the point and so there is a discontinuity for that function at x=0
good luck

steven
 

cronxeh

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Here is the graph for you to see that when you approach 0 from left you go into [tex]-\infty[/tex] and from right you get [tex]+\infty[/tex]

A unique limit doesnt exist. There is a limit when you approach from a specific side - either from positive or negative infinity, but you must specify that that is the limit you want! But a unique limit as x->0 doesnt exist because the graph diverges
 
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umm, that graph is wrong... From your graph, as x approaches infinity the limit is 0. But as x approaches infinity the limit is infinity as well... The graph resembles that of a function of the form k/x. e^x/x resembles k/x as soon as you are past x=1 and going to the negative values of x.
 

cronxeh

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That graph is correct

lim x->+inf = Inf
lim x->-inf = 0
 
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Well well, i noticed the steps. Yeah that's right but it doesn't show that x->+inf=inf.
sorry though.
 

cronxeh

Gold Member
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Well you wont see it approaching to +infinity from 0 to 1 for x obviously. You can plot it for higher values and see it grow exponentially, or you can trust the algebra and take the limit yourself
 
That graph is correct, it just has a very strange scale so it doesn't accurately portray the shape of the graph. It does show the limit x->0 very well though.
 

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