# Thread: Finding whether limit exists or not

1. ## Finding whether limit exists or not

http://i62.tinypic.com/14iol1v.png

I have watched Khanacademy videos on limit but still don't get how to solve this exercise.
Please show me step by step explanation for 2 equations in given exercise, one equation that doesn't include trigonometry and one that does.
Please link me to internet resource where I can learn how to go through this exercise. Or just let me know what concepts am I supposed to learn.

All help is appreciated, thanks.

2. ## Re: Finding whether limit exists or not

Have you not tried anything yourself? If you would show us a few where you were successful or even what you tried that did not work we would have a better idea of what kind of help you need. With, say, (m), $\displaystyle \lim_{x\to 0}x sin)(1/x)$ try letting u= 1/x. For (j) $\displaystyle \lim_{x\to\infty} \frac{x^2+ 2}{x- 5}$ a standard start is to divide both numerator and denominator by the highest power of x, here $\displaystyle x^2$. That way you wind up with "numerator/denominator" with the denominator having higher power of x. What happens to that fraction as x goes to infinity?

3. ## Re: Finding whether limit exists or not Originally Posted by AA231 Holy crap! How many of these did you expect us to help you with?? Here's a thought...Do which ones you can and ask us (no more than three at a time) to help you with the ones you don't know how to do.

-Dan

4. ## Re: Finding whether limit exists or not

For a limit to exist, when evaluating the f(x) as x approaches a from the right or left yields the same value

5. ## Re: Finding whether limit exists or not Originally Posted by AA231 http://i62.tinypic.com/14iol1v.png

I have watched Khanacademy videos on limit but still don't get how to solve this exercise.
Please show me step by step explanation for 2 equations in given exercise, one equation that doesn't include trigonometry and one that does.
Please link me to internet resource where I can learn how to go through this exercise. Or just let me know what concepts am I supposed to learn.

All help is appreciated, thanks.
I understand that you do not want answers to all these questions, just two of them. But as Halls of Ivy said, we don't know where to start because we have no idea where you are getting stuck.

A good idea when you are stuck about whether a function has a limit somewhere and if so what that limit might be is to graph the function. Does it look as if it has a limit?

$\displaystyle Given:\ g(x)\ and\ f(x)\ real\ and\ continuous\ at\ x = a\ and\ h(x) = \dfrac{f(x)}{g(x)}.$

First Rule about Limit of a Rational Function:

$\displaystyle g(a) \ne 0 \implies \lim_{x \rightarrow a}h(x) = \dfrac{f(a)}{g(a)}.$ Finite limit, no problem.

Second Rule about Limit of a Rational Function:

$\displaystyle g(a) = 0 \ne f(a) \implies |\lim_{x \rightarrow a}h(x)| = \infty.$ No finite limit, no problem.

Third Rule about Limit of a Rational Function, which a bit convoluted:

$\displaystyle g(a) = 0 = f(a),\ and\ h(x) = k(x)\ close\ to\ a,\ and\ \lim_{x \rightarrow a}k(x) \in \mathbb R \implies \lim_{x \rightarrow a}h(x) = \lim_{x \rightarrow a}k(x).$

That is, if we can find k(x) = h(x) at x close to a but not equal to a and k(x) has a limit at a then the limit for h(x) at a equals the limit for k(x). The whole trick is to figure out whether there is in fact a function k(x) that has a real limit at a and that is equivalent to h(x) close to a.

So let's take this example:

$\displaystyle h(x) = \dfrac{x^2 + x - 12}{x - 3} \implies \lim_{x \rightarrow 3}h(x) = 7\ because,\ close\ to\ 3,\ h(x) = \dfrac{x^2 + x - 12}{x - 3} = \dfrac{(x - 3)(x + 4)}{x - 3} = x + 4 = k(x)\ and\ \lim_{x \rightarrow 3}(x + 4) = 7.$

When dealing with trigonometric functions, you need to remember that $\displaystyle \lim_{x \rightarrow 0}\dfrac{sin(x)}{x} = 1 = \lim_{x \rightarrow 0}\dfrac{x}{sin(x)}.$

So for example $h(x) = \dfrac{sin(x) + 3xsin(x) }{x}.$ Does h(x) have a finite limit at x = 0?

Close to 0, $h(x) = k(x) = \dfrac{sin(x) + 3xsin(x)}{x} = = \dfrac{sin(x)}{x} + \dfrac{3xsin(x)}{x} = \dfrac{sin(x)}{x} + 3sin(x) = k(x).$

$\displaystyle \lim_{x \rightarrow 0}\left(\dfrac{sin(x)}{x} + 3sin(x)\right) = \lim_{x \rightarrow 0}\dfrac{sin(x)}{x} + 3 * \lim_{x \rightarrow 0}sin(x) = 1 + 3 * 0 = 1 \implies \lim_{x \rightarrow 0}h(x) = 1.$

6. ## Re: Finding whether limit exists or not

Hehe, can someone please show me how to do #1 and #2, I have tried doing them but came up with nothing useful.

-Thanks!

7. ## Re: Finding whether limit exists or not

What have you been smoking? (Can I have some?)

There is no "#1" and "#2"! There is only a problem 1 with 15 parts. If you mean (a) and (b), I showed you how to do (a) when you posted this same assignment on another board but I will repeat it here.

(a) There are a number of Seeing that this problem has both $\displaystyle \sqrt{x}$ and $\displaystyle \sqrt{x}$ I would let [tex]u= x^6[tex]. That way $\displaystyle \sqrt{x}= u^3$ and $\displaystyle \sqrt{x}= u^2$. That way $\displaystyle \frac{\sqrt{x}- 8}{\sqrt{x}- 4}$ becomes [tex]\frac{u^3- 8}{u^2- 4}[tex].

The first thing you should do it try setting x= 64 (or, equivalently, $\displaystyle u= 2$. If the denominator is not 0, you can just do the arithmetic. If the denominator is 0 and the numerator is not, the limit does not exist. If, as here, both are 0, we need to look more closely. It helps to know (as you should have learned in algebra) that if a polynomial is 0 when x= a, them x- a is a factor of the polynomial.

In this problem, knowing that u= 2 makes $\displaystyle u^3- 8= 0$ tells us that x- 2 is a factor. Dividing, we see that x- 2 divides into $\displaystyle x^3- 8$ $\displaystyle x^2+ 2x+ 4$ times with no remainder: $\displaystyle x^3- 8= (x- 2)(x^2+ 2x+ 4)$. Similarly the fact that x= 2 makes $\displaystyle x^2- 4= 0$ tells us that x- 2 is a factor. In fact, the fact that $\displaystyle x^2- 4= (x- 2)(x+ 2)$ is a standard product you should have learned in algebra.

So we have $\displaystyle \frac{x^3- 8}{x^2- 4}= \frac{(x- 2)(x^2+ 2x+ 4)}{(x- 2)(x+ 2)}$. The fact that we are taking the limit "as x goes to 2" means that x is NOT 0 so we x- 2 is not 0 and we can cancel the two x- 2 terms: for x not equal to 0 that fraction is the same as $\displaystyle \frac{x^2+ 2x+ 4}{x+ 2}$. Now, we can take the limit, as x goes to 2, by setting x= 2 in that expression.

(b) asks you to find the limit, as x goes to 1, of $\displaystyle \frac{1- \sqrt{x}}{1- x}$. Again, the very first thing you should try is setting x=1. You will see that doesn't work because you get "0/0" again. So do the same as above. Let $\displaystyle x= u^2$ (there is no 3 root here) so that $\displaystyle 1- \sqrt{x}= 1-u$ and $\displaystyle 1- x= 1- u^2= (1- u)(1+ u)$. Now, the fraction is $\displaystyle \frac{1- u}{(1- u)(1+ u)}$. And, as before, since x is going to 1, x, and so u, is not equal to 1, 1- u is not 0 and we can cancel the "1- u" terms so the fraction becomes $\displaystyle \frac{1}{1+ u}$. What do you get if set u= 0 in that?

8. ## Re: Finding whether limit exists or not

Thanks for the reply sorry about how I asked my previous question, I mean part a and b. What I did for a) was letting u = x^1/3 or u= x^1/2 but I couldn't get anywhere with that.
Thanks

9. ## Re: Finding whether limit exists or not

(a) Let \displaystyle \begin{align*} x = u^6 \end{align*}

(b) \displaystyle \begin{align*} 1 - x = \left( 1 - \sqrt{x} \right) \left( 1 + \sqrt{x} \right) \end{align*}

(c) same as (a)

(d) Let \displaystyle \begin{align*} x + 1 = u^{12} \end{align*}

(e)
\displaystyle \begin{align*} \frac{x}{\tan{(3x)}} &= \frac{1}{3} \cdot \frac{3x}{\tan{(3x)}} \\ &= \frac{1}{3} \cdot \frac{3x}{\frac{\sin{(3x)}}{\cos{(3x)}}} \\ &= \frac{1}{3}\cos{(3x)} \cdot \frac{3x}{\sin{(3x)}} \end{align*}

Go from here...

(f) Let \displaystyle \begin{align*} u = x - \pi \implies x = u + \pi \end{align*}, giving

\displaystyle \begin{align*} \frac{\sin{(x)}}{x - \pi} &= \frac{\sin{ \left( u + \pi \right) } }{u} \\ &= -\frac{\sin{(u)}}{u} \end{align*}

Go from here.

(g)
\displaystyle \begin{align*} \frac{\sin^2{(x)}}{x \left[ 1 - \cos{(x)} \right] } &= \frac{\sin^2{(x)}\left[ 1 + \cos{(x)} \right] }{ x \left[ 1 - \cos^2{(x)} \right] } \\ &= \frac{ \sin^2{(x)}}{ x\sin^2{(x)}} \\ &= \frac{1}{x} \end{align*}
Go from here.

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