2

Question :

Find constant numbers $a,b$ such that $\lim_{x \to 0}(x^{-3}\sin(3x)+ax^{-2}+b)=0$

My try :

I applied hopital, but it is again $\frac{0}{0}$ and the denominator becomes $x^6$ and $x^3$ which are a lot worse than the first denominator. What should i do now?

Even a hint would be appreciated. I wanna learn.

Arman Malekzadeh
  • 4,108

3 Answers3

3

Hint. One may use a Taylor series expansion, as $x \to 0$, $$ \sin x=x-\frac{x^3}{3!}+O(x^5) $$ giving $$ \frac{\sin(3x)}{x^2}=\frac{3}{x^2}-\frac{9}{2}+O(x^3). $$

Olivier Oloa
  • 120,989
2

Ok here is a hint:

$$\lim\limits_{x\to 0}\frac{x-\sin x}{x^3}=\frac{1}{6}$$

Now compaire this with your expression to get the constants.

Rene Schipperus
  • 39,526
2

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

Fudging the details a little, let's seek a solution by appealing to L'Hopital a handful of times:

$$\frac{\sin(3x)+ax+bx^3}{x^3}\to\frac{3\cos3x+a+3bx^2}{3x^2}\to\frac{-9\sin3x+6bx}{6x}\to\frac{-27\cos3x+6b}{6}$$

Now:

  • For the second of these to be of the form $0/0$, we need $3+a=0$
  • For the last of these to have limit $0$, we need $-27+6b=0$

So, $a=-3,b=27/6=9/2$ can be verified to solve the problem as stated.

For greater context, it might be helpful to consider / look at Taylor series (whether or not you're already familiar with them).

πr8
  • 10,800