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.

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

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...

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.

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.

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.

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.

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]

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