# L'Hospital's Rule

• Oct 29th 2008, 04:34 PM
Calc1stu
L'Hospital's Rule
lim -> infinite (3x+4)/(ln(4+4e^x))
lim -> 0 (2/x^4)-(5/x^2)
lim -> infinite (1+(10/x))^(x/4)
• Oct 29th 2008, 04:50 PM
Mathstud28
Quote:

Originally Posted by Calc1stu
lim -> infinite (3x+4)/(ln(4+4e^x))
lim -> 0 (2/x^4)-(5/x^2)
lim -> infinite (1+(10/x))^(x/4)

As x gets incredibely huge $4+4e^x\to{4e^x}$

So $\ln(4+4e^x)\to\ln(4)+x$

So $\frac{3x+4}{\ln(4+4e^x)}\to\frac{3x+4}{\ln(4)+x}\t o{3}$

For the secone one combining the fractions give $\frac{2-5x^2}{x^4}}$ now when you put 0 in there it should be pretty obvious what the answer is

For the third one Im sure you know that

$\lim_{x\to\infty}\left(1+\frac{1}{x}\right)^x=e$

Well in your limit if we let $\frac{10}{x}=u$ we have that

$\lim_{x\to\infty}\left(1+\frac{10}{x}\right)^{\fra c{x}{4}}=\lim_{u\to{0}}\left(1+u\right)^{\frac{10} {4u}}$

Now we can see that this is equivalent to

$\left[\lim_{u\to{0}}\left(1+u\right)^{\frac{1}{u}}\right]^{\frac{10}{4}}$

On the inner integral letting $\frac{1}{x}=u$ again we get

$\left[\lim_{x\to\infty}\left(1+\frac{1}{x}\right)\right]^{\frac{10}{4}}=e^{\frac{10}{4}}$