I have to use L'Hopital rule to figure out

lim (1+2/x)^x

x--)0

Find Partial Fraction for

-x^2-7x+27

__________

x(x^2+9)

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- Jan 17th 2006, 05:55 AMNimmyI still need help on 2 calculus problems
I have to use L'Hopital rule to figure out

lim (1+2/x)^x

x--)0

Find Partial Fraction for

-x^2-7x+27

__________

x(x^2+9) - Jan 19th 2006, 08:44 PMearboth
Hi, nimmy,

I cann't offer you a solution using the rule of de l'Hôpital, but maybe you can make some use of the following way:

In

$\displaystyle \lim_\subc{x \rightarrow 0}{\left( 1+\frac{2}{x} \right)^x}$

you can substitute $\displaystyle x=\frac{1}{n}$ because $\displaystyle \lim_{\subc{n \rightarrow \infty}}{\left(\frac{1}{n} \right) = 0}$

Your limit changes to:

$\displaystyle \lim_{\subc{n \rightarrow \infty}}{\left( 1+2n \right)^{\frac{1}{n}} = 1$

For your 2nd question I've a question too: Are you sure that you didn't make a mistake in typing the term?

Bye - Jan 19th 2006, 10:40 PMMechHead
Do you know how to do partial fraction decomposition? Not to offend, but that shouldn't really be that difficult, unless I read it wrong.

Set that whole thing equal to A/x + (Bx + C)/(x^2 + 9). Then multiply both sides by x*(x^2 + 9). You should then have

-x^2 - 7x +27 = A(x^2 + 9) + Bx^2 + Cx.

Don't ask for a proof, but I'm sure you can understand that the elements above must be grouped together according to the exponential power of x. The x^2 must be grouped together,etc. So.....

-x^2 = Ax^2 + Bx^2

-1 = A + B

-7x = Cx

C = -7

27= 9A

A = 3

Little bit of the ole algebra and yer home free!

That is how you do that, but there are other situations, like having (x-3)^2 in the denominator. Read though the strategy and I'm sure you'll get it.