# Thread: limit with l'hospitals rule

1. ## limit with l'hospitals rule

Alright. So Ill show what Ive tried to do so far and hopefully someone can help me wrap it up. So using 'hospitals rule...

lim sinxlnx
x-> 0+

so I took 2 approaches but came to a halt both ways

FIRST WAY:

lnx/(1/sinx) = FORM infin/infin

so then limit equals

= (1/x)/(-cotx cscx)

dont know where to go from here

SECOND WAY:

sinx/(1/lnx) = FORM 0/0

limit
= cosx/lnx^2

now im stuck again

Yes, this post seems messy.
So, how would I go about fully solving this problem

2. Originally Posted by chrisc
Alright. So Ill show what Ive tried to do so far and hopefully someone can help me wrap it up. So using 'hospitals rule...

lim sinxlnx
x-> 0+

so I took 2 approaches but came to a halt both ways

FIRST WAY:

lnx/(1/sinx) = FORM infin/infin

so then limit equals

= (1/x)/(-cotx cscx)

dont know where to go from here

SECOND WAY:

sinx/(1/lnx) = FORM 0/0

limit
= cosx/lnx^2

now im stuck again

Yes, this post seems messy.
So, how would I go about fully solving this problem
the trick here is to realize that close to zero, $\sin x \approx x$. thus, close to zero, $\sin x \ln x \approx x \ln x$. if you don't like that, multiply by $\frac xx$ to get $x \frac {\sin x}x \ln x$, taking the limit of the $\frac {\sin x}x$ part, we arrive at the same thing.

thus, $\lim_{x \to 0^+} \sin x \ln x = \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac {\ln x}{\frac 1x} = - \lim_{x \to 0^+} \frac {- \ln x}{\frac 1x}$

now apply L'Hopital's

3. I ended up with zero. How did I do?

4. Originally Posted by chrisc
I ended up with zero. How did I do?
yes, that's correct

here's a graph of the function, a second confirmation to your answer

5. Here is a justification what of what Jhevon did. He said $x\approx \sin x$ so we can replace them. That is how they explain it in engineering classes we can do better and explain it like they do in math classes. What Jhevon is really saying is that $\lim_{x\to 0^+}\sin x/x=1$.
That means,
$\lim_{x\to 0^+}\sin x\ln x = \lim_{x\to 0^+}\frac{\sin x}{x} \cdot x\ln x = \lim_{x\to 0^+}x\ln x$.

6. i guess ill throw this out as well
its the last question Im having trouble with, and my brain is pretty much out of energy

lim ((2x-3)/(2x+5))^(2x+1)
x->infinity

i took the ln of both sides, went to apply lhopitals rule, but no luck.
ended up getting infinity on the top, but not the bottom

7. Originally Posted by chrisc
i guess ill throw this out as well
its the last question Im having trouble with, and my brain is pretty much out of energy

lim ((2x-3)/(2x+5))^(2x+1)
x->infinity

i took the ln of both sides, went to apply lhopitals rule, but no luck.
ended up getting infinity on the top, but not the bottom
ok, let's do it your way. i still didn't end up with what you got:

Let $y = \lim_{x \to \infty} \left( \frac {2x - 3}{2x + 5} \right)^{2x + 1}$

$\Rightarrow \ln y = \lim_{x \to \infty} (2x + 1) \ln \left( \frac {2x - 3}{2x + 5} \right)$

$= \lim_{x \to \infty} \frac {\ln \left( \frac {2x - 3}{2x + 5}\right)}{\frac 1{2x + 1}} \to \frac 00$

Apply l'Hopital's

$\Rightarrow \ln y = \lim_{x \to \infty} \frac {\frac {16}{(2x + 5)^2}}{\frac {-2}{(2x + 1)^2}}$

$= - \lim_{x \to \infty} \frac {8(2x + 1)^2}{2x + 5}$

$= - \infty$

$\Rightarrow y = e^{- \infty} = 0$