
L'Hospital's Rule
I need some help solving the following problems using the L'Hospitals rule.
I'd really appreciate it if someone could walk me through the steps for solving the problem.
1. limit(x (approaches) (infinity)) xsin((pi)/x)
2. limit (x (apporaches) ((pi)/4)) (1tanx)secx
I have more but I think I'll be able to solve the rest if I find out how to do these two.

$\displaystyle \lim_{x\rightarrow\infty}\ {x \sin\ (\frac {\pi}{x}})$
When you 'substitute' $\displaystyle \infty$, you get:
$\displaystyle \lim_{x\rightarrow\infty}\ {x \sin\ (\frac {\pi}{x}})\ =\ \infty \sin (\frac {\pi}{\infty}) = \ \infty \times \sin 0 = \infty \times 0 = $ undefined.
So, we have to use L'Hopital's Rule. Differentiating, you get:
$\displaystyle \lim_{x\rightarrow\infty}\ {x \cos\ \frac{\pi}{x} \cdot \frac {\pi]{x^2}}$
Uggh, for some reason, there's a Syntax Error in my Latex Code, and I'm in a hurry, so I'll just type it out normally.
Differentiating, you get:
lim x > infinity (x . cos (pi/x) . (pi/x^2) + sin (pi/x)) > (Product Rule)
'Plugging in' infinity now gives:
(infinity. cos (pi/infinity) . (pi/infinity^2) + sin (pi/infinity))
= (infinity. cos 0 . 0 + sin 0)
= (infinity. 1. 0 + 0)
= infinity. 0 + 0
= undefined.
So, keep differentiating and 'plugging in' infinity into each derivative until you get a definite answer.
For the second one,
Plug in x = pi/4 first.
You get (1  tan (pi/4)) * sec (pi/4)
= (1  1) * (1 / cos (pi/4))
= 0 * (1 / ((sqrt (2) / 2)))
= 0 * sqrt 2
= 0.
So the limit, as x approaches pi/4, of (1  tan x) * sec x = 0.
I hope that helps.
Please forgive me if I've made any errors.
ILoveMaths07. :)

I thought you could only apply L'hospital's rule to fractions. Don't you have to convert it into one before you can calculate the answer?


My calculus professor.
Also, I checked the back of my book and the answer for the first problem should be pil

Sorry for that. I used L'Hopital's for the first one, but not for the second. The second one required just basic 'plugging and chugging'. It wasn't really L'Hopital's Rule because the function was defined for x = pi/4. The limit is 0 at x = pi/4. L'Hopital's is applied to functions which are not defined at a given point. You use it when you get 0/0 or inf/inf after plugging in the limit. And yes, you're right about L'Hopital's being applied to only fractions.
For the first one >
x sin (pi/x)
'Plugging in' infinity, we get...
infinity * sin (pi/infinity)
= infinity * sin 0
= infinity * 0
= undefined.
So you can proceed with L'Hopital's...
First, turn it into a fraction (as you mentioned):
x sin (pi/x)
= x / csc (pi/x)
Differentiating numerator an denominator separately, you get:
= 1 / (csc (pi/x) . cot (pi/x) . pi/x^2)
Now 'plug in' infinity.
You get:
= 1 / (csc (pi/infinity) . cot (pi/infinity) . pi/infinity^2)
= 1 / (csc 0 . cot 0 . 0)
= 1 / (undefined).
(Headbang)(Headbang) Same thing... (Headbang)(Headbang)

Hello,
For the first one :
substitute $\displaystyle t=\frac \pi x$
hence $\displaystyle t \to 0$ when $\displaystyle x \to + \infty$
this gives :
$\displaystyle \lim_{x \to \infty} x \sin \left(\tfrac \pi x\right)=\lim_{t \to 0} \frac \pi t \cdot \sin(t)=\pi \cdot \lim_{t \to 0} \frac{\sin(t)}{t}$
And now you can use L'Hospital's rule.

Sorry for that, again! (Headbang)(Headbang)(Headbang)
There was no way to proceed with L'Hopital's in the first place... I didn't get 0/0 or inf/inf. All I got was an undefined, even with csc, and that doesn't mean I can proceed with L'Hopital's.
Sorry... I hope you got the second one, though!
I hope that helps. :)
ILoveMaths07.

Sorry, I think what my calculus professor meant was to change the problem into a fractions so that the result is either 0/0 or (infinity)/(infinity) so we can use the L'hospital's rule.

Yes, I know what you meant. :)
ILoveMaths07.