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susus · Dec 21, 2010, 04:25 AM

lim (tanx - sinx / x^3 ) = ?
x->0

I tried to do lopital , but, I did was stuck somewhere... I got 1/4.. the

galactus · Dec 21, 2010, 06:41 AM

Is that $\lim_{x\to 0}\frac{tan(x)-sin(x)}{x^{3}}$ or what you have typed:

$\lim_{x\to 0}[tan(x)-\frac{sin(x)}{x^{3}}]$

I assume the former. Grouping symbols are important.

You can use L'Hopital, but it requires several applications.

First application:

$\frac{-cos(x)+2tan(x)sec^{2}(x)}{3x^{2}}$

second application:

$\frac{sin(x)+2tan(x)sec^{2}(x)}{6x}$

Third application:

$\frac{cos(x)+2sec^{4}(x)+4tan^{2}(x)sec^{2}(x)}{6}$

Now, letting x=0 in the numerator, we get:

$\frac{3}{6}=\frac{1}{2}$

Here is a non-L'Hopital way to go about it. I like this better.

Rewrite as:

$\lim_{x\to 0}\frac{tan(x)(1-cos(x))}{x^{3}}$

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

Now, take each limit separately.

$\lim_{x\to 0}\frac{tan(x)}{x}=\lim_{x\to 0}\left(\frac{sin(x)}{x}\right)\left(\frac{1}{cos( x)}\right)=1\cdot 1=1$

Now for:

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

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

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

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

$=(1)(\frac{1}{2})=\frac{1}{2}$

From the first limit we have 1. So, we get $(1)(1)(\frac{1}{2})$

Thus, the limit is 1/2

Note that $\lim_{x\to 0}\frac{sin(x)}{x}=1$ is a famous limit used often when proving other limits. It is not necessary to prove it here, but use it as a lemma, so to speak. It's proof usually involves the Squeeze Theorem.

As for $\lim_{x\to 0}\frac{1}{1+cos(x)}$ , we can use an approximation. Knowing that $cos(x)\approx 1-\frac{x^{2}}{2}$ for x near 0, we can sub this in and get:

$\lim_{x\to 0}\frac{2}{4-x^{2}}$ .

Now, as can be seen, as $x\to 0$ , we are left with 1/2.

Actually, we could have used this approximation, as well as $sin(x)\approx x$ for x near 0, to prove from the outset.

Doing so, leads to

$\lim_{x\to 0}\frac{tan(x)-sin(x)}{x^{3}}=\lim_{x\to 0}\frac{\frac{sin(x)}{cos(x)}-sin(x)}{x^{3}}$

$=\lim_{x\to 0}\frac{\frac{x}{1-\frac{x^{2}}{2}}-x}{x^{3}}=\lim_{x\to 0}\frac{1}{2-x^{2}}$

Now, as $x\to 0$ , we are left with... again... 1/2.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

When dealing with trig limits with $x\to 0$ , then the following approximations will prove useful.

$1-\frac{x^{2}}{2}\approx cos(x)$

$x\approx sin(x)$

$\frac{2-x^{2}}{2x}\approx tan(x)$

$\frac{2x}{2-x^{2}}\approx cot(x)$

$\frac{2}{2-x^{2}}\approx sec(x)$

$\frac{1}{x}\approx csc(x)$

These are known as 'asymptotic' approximations.