# Evaluating Limit Early

The full question is: Given function continuous from $(-1,1)$,

$f(x) = \frac{2x+p\sin{x}}{3x-2\sin{x}}, \qquad -1 < x < 0$

$f(x) = q, \qquad x=0$

$f(x) = \frac{3+2x\cot{x}}{2+x\csc{x}}, \qquad 0 < x < 1$

Find $p$, $q$

What I did was:

$\lim_{x\to 0^-} \frac{2x+p\sin{x}}{3x-2\sin{x}} = q$

$\lim_{x\to 0^-} \frac{2+p\cos{x}}{3-2\cos{x}} = 0$

$2+p = 0$

$p = -2$

Then I was stuck ... but the correct way was to use $\lim_{x\to 0} \frac{\sin{x}}{x} = \lim_{x\to 0} \frac{x}{\sin{x}} = 1$

So question is isit wrong to equate the limit to $q$ early? I got the wrong $p$ if I did that?

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What is the question? What do you need to evaluate, prove, solve? – Pedro Tamaroff Apr 17 '12 at 2:48
@PeterT.off, I updated the question. I am supposed to find $p, q$ – Jiew Meng Apr 17 '12 at 2:59
There's nothing wrong with equating to $q$, but how did the $q$ become 0 in the next line? – Ted Apr 17 '12 at 4:32
OH!, I was thinking differentiate both sides ... I should be just using L' Hopital's Rule on the Left.... – Jiew Meng Apr 17 '12 at 7:02

I'm assuming that you want to find values of $p$ and $q$ so that the function $f$ is continuous at $x=0$.

To do this, you need $$\tag{1}\lim_{x\rightarrow0^-}f(x) = f(0) =\lim_{x\rightarrow0^+}f(x).$$

You know that $f(0)=q$. You might eventually set $\lim\limits_{x\rightarrow0^-}f(x)=q$ in the solution; but you need to find the values of $p$ and $q$ so that $(1)$ holds. And to do this, you need to find the one-sided limits first.

Where do you get, from your work, that $2+p=0$? I'm not sure what you are doing here...

So, again, what you should do is evaluate the one-sided limits in $(1)$ first. You need to find

$\ \ \ \lim\limits_{x\rightarrow0^-} f(x)=\lim\limits_{x\rightarrow0^-} {2x+p\sin x\over 3x-2\sin x}$

and

$\ \ \ \lim\limits_{x\rightarrow0^+} f(x)=\lim\limits_{x\rightarrow0^+} {3+2x\cot x\over 2+x\csc x}$.

From your answer key, you have $$\lim_{x\rightarrow0^-}f(x) =2+p,\quad \text{and}\quad \lim_{x\rightarrow0^+}f(x)={5/3}.$$ Now let's see how to obtain $(1)$. The limit from the right is $5/3$ and $f(0)=q$, so $q$ is $5/3$ and that gives the equality on the right hand side of $(1)$. To get the equality on the left hand side, we set $5/3$ equal to the limit from the left: $5/3=2+p$, so $p=-1/3$.

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Thanks ... I was thinking differentiate both sides when I should be just doing it on the left ... – Jiew Meng Apr 17 '12 at 7:02

First you have to find $q$. To do this, you use the facts that $f(0)=q$, so $$q=\lim_{x\to 0^+}f(x)=\lim_{x\to 0^+}\frac{3+2x\cot x}{2+x\csc x}\;.$$

As your answer key shows, that limit is $5/3$, so $q=5/3$. Now that you know what $q$ is, you can set $$\lim_{x\to 0^-}\frac{2x+p\sin x}{3x-2\sin x}=\frac53$$ and solve for $p$. Of course to do this you have to evaluate the limit on the lefthand side; this can be done as in your answer key, so you find that $2+p=5/3$ and $p=-1/3$.

I’m not sure how you got $$\lim_{x\to 0^-}\frac{2x+p\sin x}{3x-2\sin x}=0\;,$$ unless you somehow confused $$f(x)=q\text{ when }x=0$$ with $f(0)=0$.

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Yes I made a mistake ... I was thinking differentiate both sides at the time ... – Jiew Meng Apr 17 '12 at 7:03