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)
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
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.