# I Limit with integral and absolute value

1. Oct 22, 2016

### Genilson

Hello good evening to all, I was studying here and got stuck with this.

I solved the integral and got [x+sin(x) -1]
and that´s the farthest that I got. I would appreciate the help.

2. Oct 22, 2016

### Staff: Mentor

Can you explain, what $sen$ means and where the $1$ in your antiderivative comes from?

3. Oct 22, 2016

### Staff: Mentor

I believe it is another abbreviation for sine.

4. Oct 22, 2016

### Genilson

yep , sorry (sen) = (sin).
Ok, the antiderivative should be
[ t + sin(t) ] from 0 to x.
Then I get [ x + sin(x) - 0 - sin(0) ] and Yep, I found my first mistake. sin(0) = 0

Btw I would like a solution without using L' Hopital rule, if there is one of course.

5. Oct 22, 2016

### Staff: Mentor

Are you allowed to use $\lim_{x \rightarrow 0}\frac{\sin (x)}{x}=1$? If so, you could cancel the quotient by an $x$.

6. Oct 22, 2016

### Genilson

Yep, I can use sinx/x limit but it still leave me with 0 in the denominator

7. Oct 22, 2016

### Staff: Mentor

And a $2$ in the nominator, which tells you?

Edit: Or more directly: What happens, if you divide a real number by a number that gets closer and closer to zero?

Last edited: Oct 22, 2016
8. Oct 22, 2016

### Genilson

it goes to + infinite, but Wolfram gives me as answer +inf and - inf.

9. Oct 22, 2016

### Staff: Mentor

You'd probably ignored the absolute value of your denominator. How do you manage to get a negative number?

10. Oct 23, 2016

### Genilson

I have 1 + sinx/x in the numerator, so, they´re supposed to tend to zero isn´t it? from the left or from the right it will return me 0, or am I wrong?

11. Oct 23, 2016

### Staff: Mentor

$\sin x$ and $x$ are pretty much the same in a small neighborhood around zero, aren't they? This means $\frac{\sin x}{x} \, \approx \,1$.
If $x$ is negative, then $\sin x$ is as well.

12. Oct 23, 2016

### Genilson

I've never seen it this way, thank you very much!