Slightly complicated L'Hospital's rule problem

**Bradman** · November 29th 2010, 01:07 PM

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).

**Plato** · November 29th 2010, 02:07 PM

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

**hmmmm** · November 29th 2010, 02:23 PM

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

**Plato** · November 29th 2010, 02:27 PM

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

**hmmmm** · November 29th 2010, 03:10 PM

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

**Plato** · November 29th 2010, 03:34 PM

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.

**Bradman** · November 29th 2010, 04:30 PM

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

**hmmmm** · December 2nd 2010, 02:59 AM

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

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

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

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

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

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

So we have that

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

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