MHB Limit - Values of a and b that satisifies the equation.

[Dethrone]

Hi (Wave), back already :D

For what values of $a$ and $b$ is the following equation true?

$$\lim_{{x}\to{0}}\left(\frac{\sin\left({2x}\right)}{x^3}+a+\frac{b}{x^2}\right)=0$$

I tried l'hopitals rule, but it just got more complicated.

My progress:

$$\lim_{{x}\to{0}}\left(\frac{x^2\left(\sin\left({x}\right)+ax^3+bx\right)}{x^5}\right)=0$$

For the limit to evaluate to $0$, then it has to be in indeterminate form. Hence, $\sin\left({x}\right)+ax^3+bx=0$.

Any hints?!

 

[MarkFL]

You inclination to combine terms is good, but you should have:

$$\lim_{x\to0}\left[\frac{\sin(2x)+ax^3+bx}{x^3}\right]$$

We see this in an indeterminate form of 0/0, so apply L'Hôpital's rule (remember to differentiate numerator and denominator separately). What do you get?

 

[I like Serena]

Hi (Wave), back already :D

For what values of $a$ and $b$ is the following equation true?

$$\lim_{{x}\to{0}}\left(\frac{\sin\left({2x}\right)}{x^3}+a+\frac{b}{x^2}\right)=0$$

I tried l'hopitals rule, but it just got more complicated.

My progress:

$$\lim_{{x}\to{0}}\left(\frac{x^2\left(\sin\left({x}\right)+ax^3+bx\right)}{x^5}\right)=0$$

For the limit to evaluate to $0$, then it has to be in indeterminate form. Hence, $\sin\left({x}\right)+ax^3+bx=0$.

Any hints?!

What happened to the $(2x)$ that was insided the sine?

Anyway, let's simplify to:
$$\lim_{{x}\to{0}}\left(\frac{\sin\left({2x}\right)+ax^3+bx}{x^3}\right)=0$$

This is in the indeterminate form 0/0, because:
$$\lim_{{x}\to{0}}\left(\sin\left({2x}\right)+ax^3+bx\right)
=\sin\left({2\cdot 0}\right)+a\cdot 0^3+b\cdot 0
=0$$

So... what do you get if you apply l'Hôpital's rule? (Wondering)Edit: Darn! Outmarked! ;)

 

[Dethrone]

I realized about the cancelling right when you posted, but...
$a=\frac{4}{3}$ :D

 

[I like Serena]

I realized about the cancelling right when you posted, but...
$a=\frac{4}{3}$ :D

That's quick now!
... and $b$? (Wondering)

 

[Dethrone]

I think I got it!

$$\displaystyle \lim_{x\to0}\left[\frac{\sin(2x)+ax^3+bx}{x^3}\right]=0$$

It's in indeterminate form, so the numerator must evaluate to 0.

$$\lim_{{x}\to{0}}\sin(2x)+ax^3+bx=0 $$

Knowing that $\lim_{{x}\to{0}} \frac{sin2x}{2x}=0$,

$$b=-2$$

Similarly, we can go back to the original limit and applying l'hopital's once:

$$\lim_{{x}\to{0}}\frac{2\cos\left({2x}+3ax^2+b\right)}{3x^2}=0$$

And,

$$\lim_{{x}\to{0}}2\cos\left({2x}\right)+3ax^2+b=0$$
$$\lim_{{x}\to{0}}2\cos\left({2x}\right)+4x^2+b=0$$

$$b=-2$$Finally,$$\lim_{{x}\to{0}}\frac{\sin\left({2x}\right)}{x^3}+\frac{4}{3}-\frac{2}{x^2}=0$$

Which satisfies :D

 

[I like Serena]

I think I got it!

$$\displaystyle \lim_{x\to0}\left[\frac{\sin(2x)+ax^3+bx}{x^3}\right]=0$$

It's in indeterminate form, so the numerator must evaluate to 0.

Erm... it's the other way around.
Both the numerator and the denominator evaluate to 0, so it's in indeterminate form. (Nerd)

Knowing that $\lim_{{x}\to{0}} \frac{sin2x}{2x}=0$,

Hold on! I don't know that! Is that really true? (Wondering)

Similarly, we can go back to the original limit and applying l'hopital's once:

$$\lim_{{x}\to{0}}\frac{2\cos\left({2x}+3ax^2+b\right)}{3x^2}=0$$

Hmm. Is $2\cos\left({2x}+3ax^2+b\right)$ really the derivative of $\sin(2x)+ax^3+bx$? (Wondering)

 

[MarkFL]

This is what I did...I began with:

$$\lim_{x\to0}\left[\frac{\sin(2x)+ax^3+bx}{x^3}\right]=0$$

Apply L'Hôpital:

$$\lim_{x\to0}\left[\frac{2\cos(2x)+3ax^2+b}{3x^2}\right]=0$$

In order for the value of the limit to be determinate, we must have:

$$\lim_{x\to0}\left[2\cos(2x)+3ax^2+b\right]=2+b=0\implies b=-2$$

So, we have:

$$\lim_{x\to0}\left[\frac{2\cos(2x)+3ax^2-2}{3x^2}\right]=0$$

Apply L'Hôpital and simplify:

$$\lim_{x\to0}\left[\frac{3ax-2\sin(2x)}{3x}\right]=0$$

Apply L'Hôpital:

$$\lim_{x\to0}\left[\frac{3a-4\cos(2x)}{3}\right]=0$$

$$\frac{3a-4}{3}=0$$

$$a=\frac{4}{3}$$

Hence:

$$(a,b)=\left(\frac{4}{3},-2\right)$$

 

[Dethrone]

Sorry, these were $LaTeX$ and typing errors, hehe.

About the your first statement, I wasn't clear, but I meant to say was because the denominator goes to 0, and the limit evaluates to a finite value, the numerator must also go to 0, meaning that it must be indeterminate.

2nd statement, it equals 1, don't know what I was on. (Smoking) -> too much smoking on this forum :D

Last statement, I meant $2\cos\left({2x}\right)+3ax^2+b$. I was too excited that I got the answer that I couldn't wait one second longer to submit! :D

- - - Updated - - -

This is what I did...I began with:

$$\lim_{x\to0}\left[\frac{\sin(2x)+ax^3+bx}{x^3}\right]=0$$

Apply L'Hôpital:

$$\lim_{x\to0}\left[\frac{2\cos(2x)+3ax^2+b}{3x^2}\right]=0$$

In order for the value of the limit to be determinate, we must have:

$$\lim_{x\to0}\left[2\cos(2x)+3ax^2+b\right]=2+b=0\implies b=-2$$

So, we have:

$$\lim_{x\to0}\left[\frac{2\cos(2x)+3ax^2-2}{3x^2}\right]=0$$

Apply L'Hôpital and simplify:

$$\lim_{x\to0}\left[\frac{3ax-2\sin(2x)}{3x}\right]=0$$

Apply L'Hôpital:

$$\lim_{x\to0}\left[\frac{3a-4\cos(2x)}{3}\right]=0$$

$$\frac{3a-4}{3}=0$$

$$a=\frac{4}{3}$$

Hence:

$$(a,b)=\left(\frac{4}{3},-2\right)$$

Thanks ILS and Mark!
I essentially did it this way (look at my method #2 for solving $b$), but I did it in a different order :D

 

[MarkFL]

As an aside, if you want the trademark $\LaTeX$, use the code:

\LaTeX

:D