# Limits and L'Hopital

## Main Question or Discussion Point

Hi,

I'm having trouble determining this limit:

$$\lim_{x\rightarrow0^+}\frac{\sqrt{x}\sin\sqrt{x}}{1-e^{-x}}$$​

It's a 0/0 type of expression, so it seems like L'Hopital's rule should be applicable, but no matter how many times I differentiate I keep getting 0/0-expressions. Any hints?

Also, the next question is to solve this integral:

$$\int_0^{\pi^2}\sqrt{x}\sin\sqrt{x}dx$$​

I haven't tried solving it, but since I haven't yet been able to answer any question in this whole damn exercise, I think it's a safe bet that I can't do this either. Any hints here?

TD
Homework Helper
For $\int \sqrt{x}\sin\sqrt{x}dx$

Try a substitution: $y = \sqrt x \Leftrightarrow y^2 = x \Leftrightarrow 2ydy = dx$

Which would give: $2\int {y^2 \sin \left( y \right)dy}$

For the limit, if L'Hopital doesn't work, you could do it with Taylor Series.

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when i take the derivitive of 1 - e^-x i get e^-x

and when x approches 0 i get 1 so shouldnt you get f'(x) / 1 while using l'hopitals rule
which is not 0 / 0

lurflurf
Homework Helper
$$\lim_{x\rightarrow0^+}\frac{\sqrt{x}\sin(\sqrt{x})}{ 1-e^{-x}}=\lim_{x\rightarrow0^+} \ \frac{-x}{e^{-x}-1} \ \ \frac{\sin(\sqrt{x})}{\sqrt{x}}}$$
Do these limits look familar? pehaps something like
$$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}$$
and
$$\lim_{x\rightarrow 0} \frac{e^x-1}{x}$$

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the derivative of 1 - e^(-x) at x = 0 is 0?

rachmaninoff
leright said:
the derivative of 1 - e^(-x) at x = 0 is 0?
It's 1. L'Hoptial works. So does lurflurf's suggestion, which goes back to some analysis theorems about limits of products.

Thanks for answering. But I can't see how L'Hopital works, since $$f'(x)=\frac1{2\sqrt{x}}\sin\sqrt{x} + \frac1{2}\cos\sqrt{x}$$ - you can't plug in x because of the sqrt(x) in the denominator, no?

VietDao29
Homework Helper
L'Hospital does work:
$$\lim_{x \rightarrow 0 ^ +} \frac{\sqrt{x} \sin{\sqrt{x}}}{1 - e ^ {-x}}$$
$$= \lim_{x \rightarrow 0 ^ +} \frac{\frac{\sin{\sqrt{x}}}{2 \sqrt{x}} + \frac{\cos{\sqrt{x}}}{2}}{e ^ {-x}}$$
$$= \frac{\frac{1}{2} + \frac{1}{2}}{1} = 1$$
Due to:
$$\lim_{x \rightarrow 0 ^ +} \frac{\sin{\sqrt{x}}}{\sqrt{x}} = 1$$
Viet Dao,

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lurflurf
Homework Helper
broegger said:
Thanks for answering. But I can't see how L'Hopital works, since $$f'(x)=\frac1{2\sqrt{x}}\sin\sqrt{x} + \frac1{2}\cos\sqrt{x}$$ - you can't plug in x because of the sqrt(x) in the denominator, no?
It is a limit.
What is
$$\lim_{x\rightarrow 0} \ \frac{\sin(x)}{x}$$

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I know it's a limit :D

But this is a strictly mathematical course, and I'm only supposed to use the rules in our textbook, so I can't just use rules other than these (like lim(sin(x)/x)=1, x->0). I guess, though, the way you suggest is the only possible one, so I'll do it like that anyway. Thanks!

VietDao29
Homework Helper
You can also prove:
$$\lim_{x\rightarrow 0} \ \frac{\sin(x)}{x} = 1$$
using L'Hopital rule.
$$\lim_{x\rightarrow 0} \ \frac{\sin(x)}{x} = \lim_{x \rightarrow 0}\cos{x} = 1$$
Viet Dao,

HallsofIvy
Homework Helper
If you have covered L'hopital's rule, then you have certainly already covered
$$lim_{x->0}\frac{sin x}{x}= 1$$!

(And if you haven't it is certainly simple to prove it using L'hopital's rule!)

HallsofIvy said:
If you have covered L'hopital's rule, then you have certainly already covered
$$lim_{x->0}\frac{sin x}{x}= 1$$!

(And if you haven't it is certainly simple to prove it using L'hopital's rule!)
"prove it" might not be the correct choice of words..

lurflurf
Homework Helper
broegger said:
I know it's a limit :D

But this is a strictly mathematical course, and I'm only supposed to use the rules in our textbook, so I can't just use rules other than these (like lim(sin(x)/x)=1, x->0). I guess, though, the way you suggest is the only possible one, so I'll do it like that anyway. Thanks!
I really hope
lim(sin(x)/x)=1, x->0
is in you text book
even if it were not you should be able to show it the way you show any limit.
It seems you like the theorem that for a continuous function
$$\lim_{x\rightarrow a} f(x)=f(a)$$
so here you go
let f(x)=sin(x)/x when x is not 0
and f(x)=1 when x is 0
then
lim(sin(x)/x), x->0=f(1)=1
also f can be written
$$f(x)=\int_0^1 \cos(x t) dt$$
in this form it's continuity is obvious.
or if you like power series
f(x)=1-x^2/3!+x^4/5!-x^6/7!+...+(-1)^n x^(2n)/(2n+1)!+...

rachmaninoff said:
It's 1. L'Hoptial works. So does lurflurf's suggestion, which goes back to some analysis theorems about limits of products.
I know. That's why I asked the question!

(And if you haven't it is certainly simple to prove it using L'hopital's rule!)
Is that right to say? You're assuming the derivative of $$\sin x$$ is $$\cos x$$ to prove $$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$ but actually the limit (together with the AoL) is normally used to prove the derivative of $$\sin x$$. Normally one would use the sandwich rule to find the limit.

lurflurf
Homework Helper
Gaz031 said:
Is that right to say? You're assuming the derivative of $$\sin x$$ is $$\cos x$$ to prove $$\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$$ but actually the limit (together with the AoL) is normally used to prove the derivative of $$\sin x$$. Normally one would use the sandwich rule to find the limit.
You are assuming a specific definition of sin(x). It is possible to know sin'(x)=cos(x) without knowing lim x->0 sin(x)/x=1 and the reverse. What is important is that knowing either makes it easy to find the other. What would be bad (and some books do this) would be to try to prove one of these, then get the other as a step then say it follows from what you are trying ti prove. The "Normally" you use is of course the resolution. By normally I presume you mean "How it is done in the large number of popular calculus texts that are very much alike". We need not do things as they are done normally.

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lurflurf said:
You are assuming a specific definition of sin(x). It is possible to know sin'(x)=cos(x) without knowing lim x->0 sin(x)/x=1 and the reverse. What is important is that knowing either makes it easy to find the other. What would be bad (and some books do this) would be to try to prove one of these, then get the other as a step then say it follows from what you are trying ti prove.
"Calculations" of this kind possess no conclusive power.

For example:
$$\lim_{x\rightarrow0}\frac{e^x-1}{x}$$

Differentiating the numerator and the denominator gives:
$$\lim_{x\rightarrow0}\frac{e^x}{1}=1$$

But to know that $$D(e^x) = e^x$$ we must master $$\lim_{x\rightarrow0}\frac{e^x-1}{x}$$

HallsofIvy
Homework Helper
No, that doesn't follow: as lurflurf said, there are different ways of defining sine and cosine x. For example, it is perfectly valid to define y= sin x as:
The function satisfying the differential equation y"= -y with initial value conditions
y(0)= 0, y'(0)= 1.
Then define y= cos x as:
The function satisfying the differential equation y"= -y with initial value conditions
y(0)= 1, y'(0)= 0.
That isn't normally done in basic calculus books, of course, but it is more rigorous than the usual definitions of sine and cosine functions.

It is then easy to prove that (sin x)'= cos x and then use L'hopital's rule to derive $lim_{x\rightarrow0}\frac{sin x}{x}= 1$.

Indeed, since you mention ex, many calculus texts first define ln (x) by
$ln(x)= \int_1^x \frac{1}{t}dt$. It's easy to show that that function has all the properties we expect of ln(x) including, obviously, that its derivavative is $\frac{1}{x}$ as well as the fact that it is a one-to-one function and so has an inverse.

If you then define exp(x) to be the inverse of ln(x) it follows easily that exp(x) is differentiable and has derivative equal to exp(x) itself. One could then use L'hopital's rule to show that $lim_{x\rightarrow0}\frac{e^x-1}{x}= 1$.
Once can also, by the way, prove that exp(x)= (exp(1))x.