Solving Limits Involving Rational Fractions

[iRaid]

Homework Statement

May seem easy to people but I have no idea how to do this
lim_{x\rightarrow0}\frac{sin x}{2x^{2}-x}

Homework Equations

The Attempt at a Solution

I factored out an x from the bottom but that really makes me see nothing else to do.. So I tried the quotient rule and that lead me to the wrong answer..

 

[gb7nash]

Do you know l'hopital's rule?

 

[iRaid]

Do you know l'hopital's rule?

Unfortunately no.. I've heard that's what I need for this but we haven't learned that yet..

 

[gb7nash]

Well, you don't need it for this problem, but it makes the problem much easier to solve. Did you learn taylor series yet?

 

[iRaid]

No lol this is Calc 1 basically the 2nd week of class :p

 

[gb7nash]

Maybe a better question to ask is, what have you learned with limits?

 

[iRaid]

Nvm figured it out, if anyone needs this answer PM me.

 

[HallsofIvy]

\frac{sin(x)}{2x^2- x}= \frac{sin(x)}{x(2x-1)}= \frac{sin(x)}{x}\left(2x-1\right)
and
\lim_{x\to 0}\frac{sin(x)}{x}
is a well-known limit.

 

[iRaid]

Yep that's what I ended up doing, thanks tho :D

 

[Mark44]

\frac{sin(x)}{2x^2- x}= \frac{sin(x)}{x(2x-1)}= \frac{sin(x)}{x}\left(2x-1\right)
and
\lim_{x\to 0}\frac{sin(x)}{x}
is a well-known limit.

That first line above should be

\frac{sin(x)}{2x^2- x}= \frac{sin(x)}{x(2x-1)}= \frac{sin(x)}{x}\frac{1}{2x-1}

 

[symbolipoint]

That first line above should be

\frac{sin(x)}{2x^2- x}= \frac{sin(x)}{x(2x-1)}= \frac{sin(x)}{x}\frac{1}{2x-1}

At least HallOfIvy showed his work so we could understand the method. Nice doing, both you.

 

[muzak]

Or you could have done the squeeze theorem which will be useful in your later on classes:

\frac{-1}{2x^{2}-x} \leq \frac{sin(x)}{2x^{2}-x} \leq \frac{1}{2x^{2}-x}

You can easily solve both the left and right side and see that they're both the same number, so the one in the middle has to be that number.

 

[Mark44]

Or you could have done the squeeze theorem which will be useful in your later on classes:

\frac{-1}{2x^{2}-x} \leq \frac{sin(x)}{2x^{2}-x} \leq \frac{1}{2x^{2}-x}

You can easily solve both the left and right side and see that they're both the same number, so the one in the middle has to be that number.

This technique DOES NOT apply in this problem, because both ends of the inequality are undefined at x = 0. The squeeze theorem is useful only if the two bounding functions have limits.

 

[HallsofIvy]

Or you could have done the squeeze theorem which will be useful in your later on classes:

\frac{-1}{2x^{2}-x} \leq \frac{sin(x)}{2x^{2}-x} \leq \frac{1}{2x^{2}-x}

You can easily solve both the left and right side and see that they're both the same number, so the one in the middle has to be that number.

? No, they are NOT the same number!