# Thread: Slightly complicated L'Hospital's rule problem

1. ## Slightly complicated L'Hospital's rule problem

Hi, I've been working on these problems for hours and have failed every time, so I assume there's something fundamental I'm getting wrong either in applying L'Hospital's rule or in application of derivative laws.

Here's my first problem:

Lim as x->infinity of [16x/(16x+5)]^(9x)

The answer I got was e^(45/16) which is not correct.

I'll go over how I got what I got in a bit (have to catch a bus home).

2. $\displaystyle \left[\dfrac{16x}{16x+5}\right]^{9x}=\left[1+\dfrac{-5}{16x+5}\right]^{9x}\right]\to e^{-\frac{45}{16}}$

3. do we write limit n-> infinity(1-5/(16x+5)) as e^9*(limit n->infinity(x*ln(16x/(16x+5)))

then sub in x=1/t and use l'hopitals rule to get e^9*-5(limit t->0 1/(5t+16))

which is e^-45/16

4. @hmmmm
You will find the use of LaTeX really helpful,

5. I definitely would! Especially as I'm doing a course in which I am going to have to learn it however I haven't got round to it yet, never the less would it be possible for you to tell me if my post is correct?
Thanks for the help and sorry for the inconvenience

6. Originally Posted by hmmmm
never the less would it be possible for you to tell me if my post is correct?
Well I would if I could read it. That was my point.

7. Thanks! Looks like I just lost a - sign somewhere ;-)

8. ok so i took your advice and tried my hand at some latex (it provided a good distraction from my exam revision!) and so is this correct?

we have $\displaystyle \displaystyle\lim_{x\to\infty} ( \frac{16x}{16x+5})^{9x}$

re-writing this as $\displaystyle \displaystyle\lim_{x\to\infty}exp^[\ln(( \frac{16x}{16x+5})^{9x})]$

$\displaystyle \displaystyle\lim_{x\to\infty}exp^{9x}[\ln(( \frac{16x}{16x+5}))]$

Then using the substitution $\displaystyle \frac{1}{t}$

we have $\displaystyle \displaystyle\lim_{t\to 0}exp^{\frac{9}{t}\ln\frac{\frac{16}{t}}{\frac{16} {t}+5}$

Then using L'hopitals rule on $\displaystyle \displaystyle\lim_{t\to 0}{\frac{9}{t}\ln\frac{\frac{16}{t}}{\frac{16}{t}+ 5}$=$\displaystyle \displaystyle\lim_{t\to 0}\frac{-45}{16+5t}$=$\displaystyle \frac{-45}{16}$

So we have that

$\displaystyle \displaystyle\lim_{x\to\infty} ( \frac{16x}{16x+5})^{9x}$
=$\displaystyle \displaystyle\exp^\frac{-45}{16}$