Find Limit of Integral Equation - Rules & Examples

[sara_87]

Homework Statement

are there any rules on how to find the limit of an integral equation?

for example,

find x such that the limit as y(x) tends to infinitiy of the integral equation equals 1
$$lim \int_0^x \frac{1}{y(t)-y(x)}dt=1$$

Homework Equations

The Attempt at a Solution

Im not sure how to do this, can i simply swap the limit sign with the integral sign?

Thanks in advance.

 

[gabbagabbahey]

find x such that the limit as y(x) tends to infinitiy of the integral equation equals 1
$$lim \int_0^x \frac{1}{y(t)-y(x)}dt=1$$

Huh?

$$\lim_{y(x)\to\infty}\int_0^x\frac{dt}{y(t)-y(x)}=0$$

For all values of $x$, so I'm not sure what you mean here.

 

[sara_87]

why does the limit = 0
?

what if we had t in the numerator instead of 1?

 

[jackmell]

I think you're asking:

Let:

$\lim_{x\to a} y(x)=\infty$

is there a function y(x) such that:

$\lim_{x\to a}\int_0^x \frac{1}{y(t)-y(x)}dt=1$

 

[sara_87]

no, i mean, i need to find an 'x'.

I agree that the integrand is 0 (if y tends to infinity)

but what if we have

$$\lim_{y(x)\to\infty}\int_0^x\frac{(x-t)^{-1}dt}{y(t)-y(x)}=1$$

can we find an x such that the limit is 1?

 

[Tedjn]

If you are fixing x, how can y(x) tend to infinity?

 

[sara_87]

because i need to find an x, say x* such that as y(x*) tends to infinity, the limit is 1

 

[Tedjn]

y(x*) is some number. It doesn't make sense to say that a number tends to infinity. Perhaps I'm misunderstanding the question?

 

[sara_87]

sorry, i think I am confusing things.

forget y(x*).

I want to find x such that
$$\lim_{y(x)\to\infty}\int_0^x\frac{(x-t)^{-1}dt}{y(t)-y(x)}=1$$

so y(x) is just a function depending on a variable x. then, after i found the limit (in terms of x), i want to solve the equation for x...
am i making sense?

 

[Tedjn]

Would the following interpretation be correct:

Look for x* such that

$$\lim_{x \rightarrow x^*} y(x) = \infty<br /> \quad\text{ and }\quad<br /> \lim_{x \rightarrow x^*} \int_0^x \frac{(x-t)^{-1}dt}{y(t)-y(x)} = 1,$$
​

where we always approach x* from the direction of 0.

But even if this isn't the right interpretation, it seems important to know what y(x) is as well.

 

[sara_87]

yes, this is the right interpretation.
is it possible to find x* without knowing what the function is?

 

[gabbagabbahey]

Are you sure your integrand isn't $$\frac{y(t)-y(x)}{t-x}$$ instead?

 

[sara_87]

yes, I am sure.
would it be easier if it was?

 

[Tedjn]

In addition, I don't see how to find x* without knowing the function. I mean, x* appears at a vertical asymptote, but different functions have different asymptotes.

 

[sara_87]

lets assume it has a vertical asymptote.
would we be able to find x* then?

 

[Tedjn]

Evaluate the improper integral at most at two candidates for x*, one to the left and right of zero. Then, if one of those equals 1, you're good. I don't know another way.