0
$\begingroup$

$\lim\limits_{x \to 0} f(x) = \dfrac{\sin2x}{x\cos3x}$

By the product law, can't we write:

$= 2 \cdot \lim\limits_{x \to 0} \dfrac{\sin2x}{2x} \cdot \lim\limits_{x \to 0} \dfrac{1}{\cos3x}$

Then taking the limits, replace $2x$ with $\theta$ and $\theta$ approaches $0$ if you like

$= 2 \cdot 1 \cdot \dfrac{1}{1}$

$= 2$ ?

However wolfram says it is $-\infty$ on one side and $+\infty$ on the other, and I am inclined to believe it :)

Where am I messing up?

  • 0
    You must have entered it into WA incorrectly... [See here](http://www.wolframalpha.com/input/?i=limit+sin%282x%29%2F%28x+cos%283x%29%29%2C+x+%3D0).2012-09-16
  • 1
    Huh? http://www.wolframalpha.com/input/?i=Limit%5BSin%5B2%20x%5D%2F(x%20Cos%5B3%20x%5D)%2C%20x-%3E%200%5D&t=crmtb012012-09-16
  • 0
    @DavidMitra, my apologies, I did not bracket cos(3x), I wrote cos3x, and you are right, my input was wrong. Perhaps somebody can delete this question for me.2012-09-16

1 Answers 1

1

You are correct. It appears you entered the limit into Wolfram|Alpha incorrectly, as it gives me $2$ as well here.

  • 0
    Thanks Alex, I see now that cos(3x) is necessary, cos3x is interpreted differently than I expected, I should have noticed in the nicely formatted equation wolfram presents!2012-09-16