# Math Help - [SOLVED] formal computation limits

1. ## [SOLVED] formal computation limits

I am stuck on these problems, taking the limit at x = 2.

$f(x) = \frac {sin(\pi x)} {x-2}$

$f(x) = \frac {cos \frac{\pi} {x}} {x-2}$

$f(x) = \frac {\frac{1}{x} -1} {x-2}$

Any help is greatly appreciated, including general advice about working these kinds of problems. What do you do when you can't "see" how to rationalize an equation?

2. Hi
Originally Posted by sinewave85
I am stuck on these problems, taking the limit at x = 2.

$f(x) = \frac {sin(\pi x)} {x-2}$

$f(x) = \frac {cos \frac{\pi} {x}} {x-2}$
Think about the definition of the derivative : $g'(a)=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$

For example, for the first limit : $\frac {\sin(x\pi)} {x-2}=\frac {\sin( x\pi)-\sin(2\pi)} {x-2}\underset{x\to2}{\to} ?$

3. Originally Posted by flyingsquirrel
Hi

Think about the definition of the derivative : $g'(a)=\lim_{x\to a}\frac{g(x)-g(a)}{x-a}$

For example, for the first limit : $\frac {\sin(x\pi)} {x-2}=\frac {\sin( x\pi)-\sin(2\pi)} {x-2}\underset{x\to2}{\to} ?$

I'm sorry, i'm at the very beginning of the course and they have not introduced the concept of the derivative yet. (I think that is in the next lesson.) I'm supposed to find the limit (if there is one) by using fractional reduction or by rationalizing the expression (or show that it can't be done).

4. Hello,

For the first one, substitute : $t=x-2$.

When $x \to 2$, $t \to 0$.
And $x=t+2$

---> $L=\lim_{x \to 2} \frac{\sin(\pi x)}{x-2}=\lim_{t \to 0} \frac{\sin(\pi (t+2)}{t}$

$L=\lim_{t \to 0} \frac{\sin(\pi t+2 \pi)}{t}$

But we know that for all x, $\sin(x+2 \pi)=\sin(x)$

Therefore $L=\lim_{t \to 0} \frac{\sin(\pi t)}{t}$

Substitute again, $u=\pi t$ ---> $\frac 1t=\frac{\pi}{u}$

$L=\lim_{u \to 0} \pi \frac{\sin(u)}{u}$

This is supposed to be a well-known limit :/

-----------------

If you've learnt l'Hôpital's rule, you can use it in the first step

5. Originally Posted by sinewave85
$f(x) = \frac {sin(\pi x)} {x-2}$
Using the limit composite rule this is equivalent to,
$\lim_{x\to 0} \frac{\sin [\pi(x+2)]}{x} = \lim_{x\to 0}\frac{\sin \pi x}{x} = \pi \lim_{x\to 0}\frac{\sin \pi x}{\pi x} = \pi$

Try similar things for the other ones.

6. Ok, my somewhat indifferent math education may be catching up with me. I think I understand what you did, but I am still stuck on the next problems.

$

L=\lim_{x \to 2} \frac {cos \frac{\pi} {x}} {x-2}

$

$
L=\lim_{t \to o} \frac {cos \frac{\pi} {t+2}} {t}
$

I get that I may need to use

$L=\lim_{t \to o} \frac {cos (t - 1)} {t} = 0$

but not how to get there. I have been working (unsuccessfully, obviously) at these all evening and may just be burned out, but I think I am missing something. Thanks for helping me in my hour of stupidity.

7. Originally Posted by sinewave85
Ok, my somewhat indifferent math education may be catching up with me. I think I understand what you did, but I am still stuck on the next problems.

$

L=\lim_{x \to 2} \frac {cos \frac{\pi} {x}} {x-2}

$

$
L=\lim_{t \to o} \frac {cos \frac{\pi} {t+2}} {t}
$

I get that I may need to use

$L=\lim_{t \to o} \frac {cos (t - 1)} {t} = 0$

but not how to get there. I have been working (unsuccessfully, obviously) at these all evening and may just be burned out, but I think I am missing something. Thanks for helping me in my hour of stupidity.
Seeing that

$\cos\left(\frac{\pi}{2}\right)=0$

We can see we can rewrite this as

$\lim_{x\to{2}}\frac{\cos\left(\frac{\pi}{x}\right)-cos\left(\frac{\pi}{2}\right)}{x-2}$

So seeing that this fits the definition of the derivative at a point we have

$\lim_{x\to{2}}\frac{\cos\left(\frac{\pi}{x}\right)-\cos\left(\frac{\pi}{2}\right)}{x-2}=\bigg(\cos\left(\frac{\pi}{x}\right)\bigg)'\big g|_{x=2}=\bigg(\frac{\pi\sin\left(\frac{\pi}{x}\ri ght)}{x^2}\bigg)\bigg|_{x=2}=\frac{\pi}{4}$

8. Originally Posted by sinewave85
I am stuck on these problems, taking the limit at x = 2.

$f(x) = \frac {sin(\pi x)} {x-2}$

$f(x) = \frac {cos \frac{\pi} {x}} {x-2}$

$f(x) = \frac {\frac{1}{x} -1} {x-2}$

Any help is greatly appreciated, including general advice about working these kinds of problems. What do you do when you can't "see" how to rationalize an equation?
$\lim_{x\to{2}}\frac{\frac{1}{x}-1}{x-2}\cdot\frac{x}{x}=\lim_{x\to{2}}\frac{1-x}{x^2-2x}$

Now

$1-x=-(x-1)$

and

$x^2-2x=(x-1)^2-1$

So we have

$\lim_{x\to{2}}\frac{-(x-1)}{(x-1)^2-1}$

Now let $x-1=t$

So as $x\to{2}\Rightarrow{t\to{1}}$

so we have

$\lim_{t\to{1}}\frac{-t}{t^2-1}$

Which is obviosly undefined

9. Originally Posted by Mathstud28
Seeing that

$\cos\left(\frac{\pi}{2}\right)=0$

We can see we can rewrite this as

$\lim_{x\to{2}}\frac{\cos\left(\frac{\pi}{x}\right)-cos\left(\frac{\pi}{2}\right)}{x-2}$

So seeing that this fits the definition of the derivative at a point we have

$\lim_{x\to{2}}\frac{\cos\left(\frac{\pi}{x}\right)-\cos\left(\frac{\pi}{2}\right)}{x-2}=\bigg(\cos\left(\frac{\pi}{x}\right)\bigg)'\big g|_{x=2}=\bigg(\frac{\pi\sin\left(\frac{\pi}{x}\ri ght)}{x^2}\bigg)\bigg|_{x=2}=\frac{\pi}{4}$
Ok, thanks for all of the help. But unfortunately, I am just getting more and more confused. I have not learned the derivative yet, which I realize must make this more difficult. Is there a way to get

$
L=\lim_{x \to 2} \frac {cos \frac{\pi} {x}} {x-2}
$

into the form

$
L=\lim_{x \to o} \frac {cos (x - 1)} {x} = 0
$

without using anything other than basic (precaluculus level) math? Like I said, I am at the very beginning of the course and don't know the definition of the derivative or l'Hôpital's rule. We're talking "calculus for dummies" level here -- apparently I have lost a few brain cells since the last math course I took .

10. Originally Posted by sinewave85
Ok, thanks for all of the help. But unfortunately, I am just getting more and more confused. I have not learned the derivative yet, which I realize must make this more difficult. Is there a way to get

$
L=\lim_{x \to 2} \frac {cos \frac{\pi} {x}} {x-2}
$

into the form

$
L=\lim_{x \to o} \frac {cos (x - 1)} {x} = 0
$

without using anything other than basic (precaluculus level) math? Like I said, I am at the very beginning of the course and don't know the definition of the derivative or l'Hôpital's rule. We're talking "calculus for dummies" level here -- apparently I have lost a few brain cells since the last math course I took .
FYI

$\lim_{x\to{0}}\frac{\cos(x-1)}{x}\ne{0}$

But

$\lim_{x\to{0}}\frac{1-\cos(x)}{x}=0\Rightarrow-\lim_{x\to{0}}\frac{\cos(x)-1}{x}=-0=0$

But hmm,

I would suggest making a substitution that lets you expand the cosine by trig identites

11. Originally Posted by Mathstud28
$\lim_{x\to{2}}\frac{\frac{1}{x}-1}{x-2}\cdot\frac{x}{x}=\lim_{x\to{2}}\frac{1-x}{x^2-2x}$

Now

$1-x=-(x-1)$

and

$x^2-2x=(x-1)^2-1$

So we have

$\lim_{x\to{2}}\frac{-(x-1)}{(x-1)^2-1}$

Now let $x-1=t$

So as $x\to{2}\Rightarrow{t\to{1}}$

so we have

$\lim_{t\to{1}}\frac{-t}{t^2-1}$

Which is obviosly undefined
Thanks for the help on this one. I did most of the things you did, never got to that last step of putting it into a form that proves that it does not exist. I kept trying to make it exist, which apparently was the wrong tact!

12. Originally Posted by sinewave85
Thanks for the help on this one. I did most of the things you did, never got to that last step of putting it into a form that proves that it does not exist. I kept trying to make it exist, which apparently was the wrong tact!
You do know that for

$\lim_{x\to{c}}f(x)$ to exist we must have that

$\lim_{x\to{c^+}}f(x)=\lim_{x\to{c^{-}}}f(x)$

Right? Or in other words for a limit to exist we must have that the limit from the left equals the limit from the right

So in this case

$\lim_{x\to{1^-}}\frac{-x}{x^2-1}=\infty$

whereas

$\lim_{x\to{1^+}}\frac{-x}{x^2-1}=-\infty$

$\therefore\quad\lim_{x\to{1}}\frac{-x}{x^2-1}\text{ Does not exist}$

13. Ok, I have got it now. Thanks for the help, all.