Limit with integral and absolute value

  • I
  • Thread starter Genilson
  • Start date
  • #1
6
0
Hello good evening to all, I was studying here and got stuck with this.
upload_2016-10-22_17-23-36.png

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

Answers and Replies

  • #2
14,169
11,470
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?
 
  • Like
Likes Genilson
  • #4
6
0
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
14,169
11,470
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##.
 
  • Like
Likes Genilson
  • #6
6
0
Yep, I can use sinx/x limit but it still leave me with 0 in the denominator
 
  • #7
14,169
11,470
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?
 
Last edited:
  • Like
Likes Genilson
  • #8
6
0
it goes to + infinite, but Wolfram gives me as answer +inf and - inf.
 
  • #9
14,169
11,470
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?
 
  • Like
Likes Genilson
  • #10
6
0
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
14,169
11,470
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.
 
  • Like
Likes Genilson
  • #12
6
0
I've never seen it this way, thank you very much!
 

Related Threads on Limit with integral and absolute value

  • Last Post
Replies
2
Views
8K
Replies
3
Views
2K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
2
Views
5K
  • Last Post
Replies
2
Views
676
  • Last Post
Replies
3
Views
4K
  • Last Post
Replies
3
Views
13K
  • Last Post
Replies
4
Views
19K
Replies
3
Views
5K
Replies
2
Views
2K
Top