Finding limit of $\frac{\tan{x}-\sin{x}}{x^3}$.

Question

I recently got this question in my yearly mathematics exam and was one of the questions I calculated wrong.

What I did was :- $$\lim_{x\to 0}{\frac{\tan{x}-\sin{x}}{x^3}}$$ $$=\lim_{x\to 0}{\frac{\tan{x}}{x^3}}-{\frac{\sin{x}}{x^3}}$$ $$=\lim_{x\to 0}{\frac{\to 0}{x^2}}-{\frac{\to 0}{x^2}}$$ $$=\lim_{x\to 0}{(\to 0)({\frac{1}{x^2}}-{\frac{1}{x^2}})}=0$$ which is wrong. The correct answer is $\frac{1}{2}$ as given by wolfram alpha.

So my question is :-

1) What's the error in my answer?

2) How is the correct answer to be calculated?

Thanks for help :)

Answer 1

The following step in you calculation was wrong:

$$=\lim_{x\to 0}{\frac{\tan{x}}{x^3}}-{\frac{\sin{x}}{x^3}}$$ $$=\lim_{x\to 0}{\frac{\to 0}{x^2}}-{\frac{\to 0}{x^2}}$$

You didn't write it down, but in order to do this you need to do an additional step, that is:

$$\lim_{x\to 0}{\frac{\tan{x}}{x^3}}-{\frac{\sin{x}}{x^3}}$$ $$=\lim_{x\to 0}{\frac{\tan{x}}{x^3}}-\lim_{x\to 0}{\frac{\sin{x}}{x^3}}.$$

But the above is not true. Separating sums in limits is only valid, if the limits of both summands exist and are finite, but in this case they don't.

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A valid calculation of the limit would for example be:

\begin{align} &\lim_{x\to 0}{\frac{\tan{x}-\sin x}{x^3}} \\ =& \lim_{x\to 0} \frac {\tan x}x \frac{1-\cos x}{x^2} \\ =& \lim_{x\to 0} \frac {\tan x}x \lim_{x\to 0}\frac{1-\cos x}{x^2} \\=&1\cdot \frac 1 2 =\frac 1 2. \end{align} Here we can calculate each limit separately and multiply them afterwards, as both exist.
(In the first step I just took the $\tan x$ factor outside).

Answer 2

Hint: Use the Taylor development: $sin(x)=x-x^3/6+O(x^3)$, $tan(x)=x+x^3/3+O(x^3)$.

Answer 3

we have $$\frac{-\sin(x){\cos(x)}+\sin(x)}{\cos(x)x^3}=\frac{\sin(x)(1-\cos^2(x))}{x^3\cos(x)(1+\cos(x))}=\frac{\sin^3(x)}{x^3}\frac{1}{\cos(x)(1+\cos(x))}$$