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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...

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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...

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[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.

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[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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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

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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.

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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

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So what's wrong with infinity? This limit does not exist. The function elwrthy 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???

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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

Aaaahh I see my mistake there.

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

[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

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

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

steven

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Thus your answer's wrong steven... Or am I doing something wrong?

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[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

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If a functionwisredz said:I still cannot understand, a function cannot have two limits at a point, and this one has.

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I think your friends set you up dude

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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.

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No, you are right. There is no limit. the limit as x->0 of e^x/x is undefined.wisredz said:

Thus your answer's wrong steven... Or am I doing something wrong?

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[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

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

[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

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steven187 said:

[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?

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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

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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

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

Last edited:

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That graph is correct

lim x->+inf = Inf

lim x->-inf = 0

lim x->+inf = Inf

lim x->-inf = 0

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sorry though.

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