l'Hospital's rule help

[dvdman2k]

Recently the class professor went over l'Hospital's rule but didn't go into as much detail as I would have liked and was wondering if possible someone could inform me as to how you would do problems such as these.

1. Find lim goes to 0 (zero) ((e^11x)-1-11x)/x^2 (would that be done by taking the derivative of the top and bottom and using the derivative division rule something like:
f= ((e^11x)-1-11x)
f'=((e^11x)-11)
g=(x^2)
g'=2x
then doing (g*f'-f*g')/g^2?
2. Find lim x goes to infiniti 6x/ln(7+5e^x) (I think you use the chain rule along with the division rule?)
3. Find lim goes to 0 (zero) x/tan^-1*(2x)
4. lim x goes to infiniti x[ln(x+9)-lnx]
Thank you.

 

[galactus]

I'll go ahead and step thorugh the first one. You try the rest. Okey-doke?.

Yes, you take the derivatives of the den and num, respectively.

DO NOT use the quotient rule.

\(\displaystyle \L\\\lim\frac{f'(x)}{g'(x)}\)

\(\displaystyle \L\\\lim_{x\to\0}\frac{e^{11x}-1-11x}{x^{2}}\)

\(\displaystyle \L\\f'(x)=11e^{11x}-11\)

\(\displaystyle \L\\g'(x)=2x\)

where \(\displaystyle \L\\\frac{f'(x)}{g'(x)}=\frac{11e^{11x}-11}{2x}\)

Take the derivatives twice(yes, you can do that) and get:

\(\displaystyle \L\\\lim_{x\to\0}\frac{121e^{11x}}{2}=\frac{121}{2}\)