# Limit with integral and absolute value

• I
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.

fresh_42
Mentor
Hello good evening to all, I was studying here and got stuck with this.
View attachment 107864
I solved the integral and got [x+sin(x) -1]
and that´s the farthest that I got. I would appreciate the help.
Can you explain, what ##sen## means and where the ##1## in your antiderivative comes from?

Genilson
Mark44
Mentor
Can you explain, what sen means
I believe it is another abbreviation for sine.

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.

fresh_42
Mentor
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.
Are you allowed to use ##\lim_{x \rightarrow 0}\frac{\sin (x)}{x}=1##? If so, you could cancel the quotient by an ##x##.

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

fresh_42
Mentor
Yep, I can use sinx/x limit but it still leave me with 0 in the denominator
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?

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

fresh_42
Mentor
it goes to + infinite, but Wolfram gives me as answer +inf and - inf.
You'd probably ignored the absolute value of your denominator. How do you manage to get a negative number?

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?

fresh_42
Mentor
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?
##\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.

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