Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $\lim_{x\to 0} \frac{\sin{x}}{x} = \lim_{x\to 0} \frac{x}{\sin{x}} = 1$

enter image description here

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

share|cite|improve this question
What is the question? What do you need to evaluate, prove, solve? – Pedro Tamaroff Apr 17 '12 at 2:48, 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
up vote 5 down vote accepted

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} $


$\ \ \ \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 $p=-1/3$.

share|cite|improve this answer
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 $f(0)=0$.

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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.