Calc 1, limit error

Question

Apparently the following equals 2, what I'm getting though is 0.

Where is the mistake?

Find: $ \lim\limits_{x\rightarrow0} \frac{x^2}{1 - cosx} $

$ \lim\limits_{x\rightarrow0} \frac{x*x}{1 - cosx} $ (expand numerator)

$ \lim\limits_{x\rightarrow0} x*\frac{x}{1 - cosx} $ (move x outside numerator)

$ \lim\limits_{x\rightarrow0} x $ * $ \lim\limits_{x\rightarrow0}\frac{x}{1 - cosx} $ (product property of limits)

$ \lim\limits_{x\rightarrow0} x $ * $ \lim\limits_{x\rightarrow0}\left(\frac{1 - cosx}{x}\right)^{-1} $ (inverse the fraction of the second limit)

$ \lim\limits_{x\rightarrow0} x = 0 $ , and we know also that $ \lim\limits_{x\rightarrow0}\left(\frac{1 - cosx}{x}\right)=0 $

Therefore:

$ 0 * 0^{-1} $

$=0$

Answer 1

First of all, dividing by zero is a tricky business; statements like $0 \cdot 0^{-1} = 0$ make a mathematician shiver.

So the first step would be acknowledging that we are dealing with an indeterminacy; therefore, we should try to get rid of it.

Applying L'Hôpital's rule twice we get:

$$ \lim_{x \to 0} \frac{x^2}{1 - \cos x} = \lim_{x \to 0} \frac{2x}{\sin x} = \lim_{x \to 0} \frac{2}{\cos x} = 2 $$

Without L'Hôpital's rule, and remembering that $\lim_{x \to 0} \frac{sin x}{x} = 1$, we can do something like:

$$ \lim_{x \to 0} \frac{x^2}{1 - \cos x} = \lim_{x \to 0} \frac{x^2}{2 \sin^2 (\frac{x}{2})} = \lim_{x \to 0} 2 \cdot \frac{(\frac{x}{2})^2}{sin^2 \frac{x}{2}} = \lim_{x \to 0} 2 \cdot (\frac{\frac{x}{2}}{sin \frac{x}{2}})^2 = \lim_{x \to 0} 2 \cdot (\frac{sin \frac{x}{2}}{\frac{x}{2}})^{-2} = \lim_{x \to 0} 2 \cdot 1^{-2} = 2 $$

Answer 2

With the power series for $ \cos$:

$$\frac{1- \cos x}{x^2}= \frac{1}{2}-\frac{x^2}{4!}+\frac{x^4}{6!}- +.... \to \frac{1}{2}$$

as $x \to 0.$ Hence

$$\lim\limits_{x\rightarrow0} \frac{x^2}{1 - \cos x}=2.$$

Answer 3

I believe this is a Calc 1 appropriate answer.

We have \begin{align} \lim_{x\to0}\frac{x^2}{1-\cos x}&=\lim_{x\to0}\frac{x^2}{1-\cos x}\cdot\frac{1+\cos x}{1+\cos x}\\ &=\lim_{x\to0}\frac{x^2(1+\cos x)}{1-\cos^2x}\\ &=\lim_{x\to0}\frac{x^2(1+\cos x)}{\sin^2x}\\ &=\lim_{x\to0}\frac{x^2}{\sin^2x}\cdot(1+\cos x)\\ &=\lim_{x\to0}\frac{x^2}{\sin^2x}\cdot\lim_{x\to0}(1+\cos x)\\ &=1\cdot2\\ &=\boxed2 \end{align} as desired.