Anyone wanna help me solve this limit? :p

[T.Vengeance]

I've been on this for half an hour...

[Sklo]

^3 is the problem, isn't there a preset formula for that?(was some time ago)

[T.Vengeance]

^3 is the problem, isn't there a preset formula for that?(was some time ago)

It's an infinite limit problem that's why. Idk how to solve it cause I can't cancel out the bottom part.

 

Nevermind. Got it.

[kichilron]

I don't understand? Maybe I am missing something, but it looks quite easy to me.

 

If x -> 5, then e^x is constant.

 

And x-5 will be 0, meaning it will be:

 

(e⁵) / 0³

 

This is allowed, as it's limes and this means the result is: infinite

[joncppl]

This will really help in cases like these (and much more complicated limits as well, so learn it)
http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

Rule of thumb, to quickly tell: power off numerator > power of denominator, therefore diverge to +infinty

[Sklo]

I don't understand? Maybe I am missing something, but it looks quite easy to me.

 

If x -> 5, then e^x is constant.

 

And x-5 will be 0, meaning it will be:

 

(e⁵) / 0³

 

This is allowed, as it's limes and this means the result is: infinite

he is right, just figured it out again.

[TideRower]

I don't understand? Maybe I am missing something, but it looks quite easy to me.

 

If x -> 5, then e^x is constant.

 

And x-5 will be 0, meaning it will be:

 

(e⁵) / 0³

 

This is allowed, as it's limes and this means the result is: infinite

You forget that we have x -> 5^-, which means the denominator is negative. Therefor the result is -infinite.

 

This will really help in cases like these (and much more complicated limits as well, so learn it)

http://en.wikipedia.org/wiki/L'H%C3%B4pital's_rule

Rule of thumb, to quickly tell: power off numerator > power of denominator, therefore diverge to +infinty

Your rule of thumb doesn't work in this case.

[kichilron]

You forget that we have x -> 5^-, which means the denominator is negative. Therefor the result is -infinite.

 

You are very correct, even though around here it means - when not otherwise stated - that lim x->5 == lim x->5+

[TideRower]

You are very correct, even though around here it means - when not otherwise stated - that lim x->5 == lim x->5+

 

That doesn't seem to make a lot of sense, since 1 is also equal to +1. Also +infinite is used more frequently than -infinite (not talking about limts).

Your statement about the limits being equal, is strange. If I write nothing I mean that the limes is coming from both sides.

We have: lim x->5 <=> lim x->5^-  and lim x->5⁺

[TheGosuStandard]

Wouldn't it be approaching infinity from the left as the denominator gets ever so close to getting to zero, where as the numerator will get larger & larger?

 

Found someone from yahoo that answered this: 

 

"(lim x-->5-) e^x/(x - 5)^3 

Plugging in x = 5 gives us: 

e^5/(5 - 5)^3 
= e^5/0 

Now, through negative values, this function will approach negative values as (x - 5)^3 < 0 for x < 5. 

Therefore, this approaches negative infinity."

[kichilron]

That doesn't seem to make a lot of sense, since 1 is also equal to +1. Also +infinite is used more frequently than -infinite (not talking about limts).

Your statement about the limits being equal, is strange. If I write nothing I mean that the limes is coming from both sides.

We have: lim x->5 <=> lim x->5^-  and lim x->5⁺

 

 

Wisest would most likely be to just say:

 

+- infinite

[TideRower]

Wisest would most likely be to just say:

 

+- infinite

 

If we had, lim x->5 I would write +-infinite. Otherwise the differencation between + and - infinite is important.

There is however a case where no difference exists.