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Here is a homework problem I am having trouble with:

If $$f(x) = \frac{\sin{3x} +a\sin{2x} + b\cos{x}}{x^3}$$

is continuous at $x=0$, find the values of $a$ and $b$.

I noticed I cannot apply L'Hopital's rule as the numerator doesn't diminish at $x=0$. I have no idea how to take limit for this function at $x = 0$

Any hint or suggestion would be appreciated.

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  • $\begingroup$ "as the numerator doesn't diminish at $x=0$" Well, it must if $f$ is to be continuous. That gives a condition on $b$. Then you need to find a condition on $a$. $\endgroup$ Commented Aug 11, 2013 at 16:21
  • $\begingroup$ @Daniel: But $f$ is continuous everywhere it is defined. It doesn't matter what $a$ and $b$ are. $\endgroup$ Commented Aug 12, 2013 at 2:25

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Hint

Why $b$ should be $0$? and use the Taylor series of $\sin$: $$\sin x=x-\frac{x^3}{6}+O(x^5)$$

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Hint: If $b\ne 0,$ then indeed, the numerator does not vanish as $x\to 0,$ while the denominator does, and so $\lim_{x\to 0}f(x)$ does not exist. Now, put $b=0$ and see if you can find $a$.

As a side note, the function is not defined at $x=0$, so not continuous there, regardless of the values of $a$ and $b$. You're trying to find $a,b$ so that $\lim_{x\to 0}f(x)$ exists--equivalently, so that $f(x)$ can be extended to a function continuous everywhere.

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HINT:

As $f(x)$ is continuous at $x=0,$ and $$\lim_{x\to0}f(x)=\frac b0$$ must be defined & finite $\implies b=0 $

$\implies f(x)=\sin3x+a\sin2x$

As $$\lim_{x\to0}\frac{f(x)}{x^3}\text{ is of the form }\frac00$$

we can apply L'Hospital's Rule here

$$\implies \lim_{x\to0}\frac{f(x)}{x^3}=\lim_{x\to0}\frac{\sin3x+a\sin2x}{x^3}=\lim_{x\to0}\frac{3\cos3x+2a\cos2x}{3x^2}=\frac{3+2a}0$$

so the numerator must be $0$ to keep the limit finite & defined

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  • $\begingroup$ No, $f(x)$ is not continuous at $x=0,$ regardless of our choice of $a$ and $b$. It isn't defined there at all. $\endgroup$ Commented Aug 12, 2013 at 2:27

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