Limits Question

[BobbyJoeCool]

Ok, I'm taking Calculus I. My teacher is not very good (he's a grad student and not very good at teaching. He knows what he's talking about, but not good at explaining).

 

This was his problem:

 

[math]\lim_{n\to 8} \frac{e^x-e^8}{\sqrt{x+1}-3}[/math]

 

Now, direct substitution says

 

[math]\frac{e^8-e^8}{\sqrt{8+1}-3}=\frac{0}{\sqrt{9}-3}=\frac{0}{3-3}=\frac{0}{0}[/math]

 

So, we tried to get the 0 out of the denominator bu multiplying by the congigate.

 

[math]\frac{(e^x-e^8)(\sqrt{x+1}+3)}{(\sqrt{x+1}-3)(\sqrt{x+1}+3)}[/math]

[math]\frac{(e^x-e^8)(\sqrt{x+1}+3)}{x+1-9}[/math]

[math]\frac{(e^x-e^8)(\sqrt{x+1}+3)}{x-8}[/math]

 

So try direct subsitiution again...

 

[math]\frac{(e^8-e^8)(\sqrt{8+1}+3)}{8-8}[/math]

[math]\frac{(0)(\sqrt{9}+3)}{0}=\frac{0*(3+3)}{0}=\frac{0}{0}[/math]

 

I know the limit exists. Furthermore, I know the limit is [math]6e^8[/math] because I put it into my TI-89 calulator (which he won't let us use on tests) and that is what it gave me, but we couldn't find it.

 

Something seemed a little odd to me about the x-8 in the denominator and the exponents on the "e"'s being x and 8, so I thought it might have something to do with that. But can anyone help me figure this out? Knowing me, it's probably something very simple, and I just don't see it, but I'd like to know.

[Bubba]

This is a perfect examples of where to use l'Hopitals rule, if you'v learnt it yet, otherwise i can't help sorry.

[BobbyJoeCool]

This is a perfect examples of where to use l'Hopitals rule, if you'v learnt it yet, otherwise i can't help sorry.

 

sounds farmiliar, but I can't seem to remember it right now...

[mezarashi]

L'Hopital's Rule: Basically you differentiate the numerator term and differentiate the denominator term SEPARATELY. Then do substitution again. If you still get the indeterminate form of 0/0 or infinity/infinity, then you may repeat the rule again.

[BobbyJoeCool]

ahhhh...

 

so

 

[math]\frac{\frac{dx}{dy}((e^x-e^8)(\sqrt{x+1}+3))}{\frac{dx}{dy}(x-8)}[/math]

 

as the rule, where u and v are both variables [math]\frac{dx}{dy}u*v=u*v'+v*u'[/math]

 

[math]\frac{dx}{dy}((e^x-e^8)(\sqrt{x+1}+3))[/math]

 

[math](e^x-e^8)\frac{dx}{dy}(\sqrt{x+1}+3)+(\sqrt{x+1}+3)\frac{dx}{dy}(e^x-e^8)[/math]

 

Start with the easy one...

 

[math]\frac{dx}{dy}(e^x-e^8)=e^x[/math]

 

now the harder one...

 

[math]\frac{dx}{dy}(\sqrt{x+1}+3)[/math]

 

[math]1/2*(x+1)^{-1/2}=\frac{1}{2\sqrt{x+1}}[/math]

 

backsubstitute

 

[math](e^x-e^8)\frac{1}{2\sqrt{x+1}}+(\sqrt{x+1}+3)e^x[/math]

 

[math]\frac{(e^x-e^8)}{2\sqrt{x+1}}+\frac{(\sqrt{x+1}+3)e^x}{1}[/math]

 

[math]\frac{(e^x-e^8)}{2\sqrt{x+1}}+\frac{2\sqrt{x+1}(\sqrt{x+1}+3)e^x}{2\sqrt{x+1}}[/math]

 

[math]\frac{(e^x-e^8)+2\sqrt{x+1}(\sqrt{x+1}+3)e^x}{2\sqrt{x+1}}[/math]

 

[math]\frac{e^x-e^8+(2(x+1)+6\sqrt{x+1})e^x}{2\sqrt{x+1}}[/math]

 

[math]\frac{e^x-e^8+(2x+2+6\sqrt{x+1})e^x}{2\sqrt{x+1}}[/math]

 

[math]\frac{((2x+2)+6\sqrt{x+1})+1)e^x-e^8}{2\sqrt{x+1}}[/math]

 

[math]\frac{(6 \sqrt{x+1}+2x+3)e^x-e^8}{2\sqrt{x+1}}[/math]

 

and

 

[math]\frac{dx}{dy}(x-8)=1[/math]

 

so

 

[math]\frac{\frac{(6 \sqrt{x+1}+2x+3)e^x-e^8}{2 \sqrt{x+1}}}{1}[/math]

 

Therefor, we substitute x=8

 

[math]\frac{\frac{(6 \sqrt{8+1}+2*8+3)e^8-e^8}{2 \sqrt{8+1}}}{1}[/math]

 

[math]\frac{(6 \sqrt{9}+16+3)e^8-e^8}{2 \sqrt{9}}[/math]

 

[math]\frac{(6*3+16+3)e^8-e^8}{2*3}[/math]

 

[math]\frac{(18+16+3)e^8-e^8}{6}[/math]

 

[math]\frac{(37)e^8-e^8}{6}[/math]

 

[math]\frac{36e^8}{6}[/math]

 

[math]\frac{36}{6}e^8[/math]

 

[math]6e^8[/math]

 

Right?

 

And my teacher chose to choose this as a probelm to give us whilst "learning" limits before learning how to diferentiate... I see.

[mezarashi]

Seems like it. Simple test, plug in 7.999999 or 8.000001, and see what you get

 

It's a bit strange that your teacher would give you this problem before teaching how to differentiate in the first place but oh well... maybe there's another way to do it.

[oatjay]

Seems like it. Simple test' date=' plug in 7.999999 or 8.000001, and see what you get

[/quote']

 

 

i don't think the 7.9999 thing would work very well in this case... it would take a bit of a leap i think to go from 17885.748 to 6e^8. that's why it helps to know how to do these things manually sometimes.

[PhDP]

Even simpler...

 

[math]

\lim_{x\to 8} \frac{e^x-e^8}{\sqrt{x+1}-3}

[/math]

 

With L'Hospital's Rule;

 

[math]

\lim_{x\to 8} \frac{\frac{d}{dx}\left(e^x-e^8\right)}{\frac{d}{dx}\left(\sqrt{x+1}-3\right)}

[/math]

 

[math]

\lim_{x\to 8} \frac{\frac{d(e^x)}{dx}-\frac{d(e^8)}{dx}}{\frac{d(\sqrt{x+1})}{dx}-\frac{d(3)}{dx}}

[/math]

 

[math]\lim_{x\to 8} \frac{e^x}{\frac{1}{2\sqrt{x+1}}}[/math]

 

[math]e^8 \times 2\sqrt{8+1}}}[/math]

 

[math]6e^8[/math]

[BobbyJoeCool]

good point on that one. I just hate taking the derivative of a nth root. *shudders*