# Thread: Limit as x->0 of a complicated trig function

1. ## Limit as x->0 of a complicated trig function

The function is:

(tan(x))^3 - x
x+x^2
Allowed facts to use are that the limit as x tends to 0 of sin(x)/x=1, and theorems including the algebra of limits, sandwich theorem and the 'changing the variables' theorem.

I believe the limit to be -1, and I'm confident of showing it using the thorems stated above, so my problem isn't specifically with showing the limit, it's more with formatting the problem. Re-writing the initial function in such a way that I can use the limit of sin(x)/x.

I started by taking (tanx)^3= (sin(x)^3)/(cos(x)^3), and then (cos(x))^3= cos(x)(1-(sin(x))^2). But to be honest it all gets rather complex, and as I haven't learnt Latex yet it would look an absolute mess on here, and also my trigonometry is a bit rusty, so if anyone could help me through this initial part, or offer an easier method, I'd be grateful. Thanks!

2. ## Re: Limit as x->0 of a complicated trig function

Originally Posted by struggler
The function is:

(tan(x))^3 - x
x+x^2
Allowed facts to use are that the limit as x tends to 0 of sin(x)/x=1, and theorems including the algebra of limits, sandwich theorem and the 'changing the variables' theorem.

I believe the limit to be -1, and I'm confident of showing it using the theorems stated above, so my problem isn't specifically with showing the limit, it's more with formatting the problem. Re-writing the initial function in such a way that I can use the limit of sin(x)/x.

I started by taking (tanx)^3= (sin(x)^3)/(cos(x)^3), and then (cos(x))^3= cos(x)(1-(sin(x))^2). But to be honest it all gets rather complex, and as I haven't learned Latex yet it would look an absolute mess on here, and also my trigonometry is a bit rusty, so if anyone could help me through this initial part, or offer an easier method, I'd be grateful. Thanks!
Yes, the limit is -1.

Probably the easiest way to show this is with Taylor series. I suppose the is not allowed for you.

After taking (tanx)^3= (\sin(x)^3)/(\cos(x)^3), multiply the whole expression by $\frac{\displaystyle \frac{\cos(x)}{x}}{\displaystyle \frac{\cos(x)}{x}}\ .$

3. ## Re: Limit as x->0 of a complicated trig function

Equivalently, write $tan^3(x)= \frac{sin^3(x)}{cos^3(x)}$ and $x^2+1= x(x+1)$. Now you can write the fraction as
$\frac{sin(x)}{x}\frac{sin^2(x)}{cos^3(x)(x+1)}- \frac{1}{x^2+1}$
and take the limit of each fraction separately.

4. ## Re: Limit as x->0 of a complicated trig function

You need to know that:

$\lim_{x \to 0} \frac{\tan(x)}{x}=1$

which you can derive from the equivalent limit with $\sin(x)$ in place of $\tan(x)$.

Then:

$\lim_{x \to 0} \frac{\tan^3(x)-x}{x+x^2}=\lim_{x \to 0} \frac{\frac{\tan^3(x)}{x^3}x^3-x}{x+x^2}=\lim_{x \to 0} \frac{\frac{\tan^3(x)}{x^3}x^2-1}{1+x}=-1$

CB (posting a full solution because after 10 days off line this can no longer be handed in for credit)

5. ## Re: Limit as x->0 of a complicated trig function

There is another way to do it.

It is known that that as $x\to 0$, $\sin{x}\approx x$ and $\cos{x}\approx 1-\frac{x^2}{2}$.

Limit becomes: $\\ \lim_{x\to 0}\frac{\tan^3{x}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{\sin^3x}{\cos^3x}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{8x^3}{(2-x^2)^3}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{8x^2}{(2-x^2)^3}-1}{x+1}=-1$

6. ## Re: Limit as x->0 of a complicated trig function

Originally Posted by sbhatnagar
There is another way to do it.

It is known that that as $x\to 0$, $\sin{x}\approx x$ and $\cos{x}\approx 1-\frac{x^2}{2}$.

Limit becomes: $\\ \lim_{x\to 0}\frac{\tan^3{x}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{\sin^3x}{\cos^3x}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{8x^3}{(2-x^2)^3}-x}{x+x^2}=\lim_{x\to 0}\frac{\frac{8x^2}{(2-x^2)^3}-1}{x+1}=-1$
Without making precise what you mean by " $\approx$" this does not work as it is. My post uses this idea, but in the form of the limit of tan(x) over x, which does work without further ado.

CB

7. ## Re: Limit as x->0 of a complicated trig function

Thanks for all your advice everyone, I used something along these lines in the end
Originally Posted by HallsofIvy
Equivalently, write $tan^3(x)= \frac{sin^3(x)}{cos^3(x)}$ and $x^2+1= x(x+1)$. Now you can write the fraction as
$\frac{sin(x)}{x}\frac{sin^2(x)}{cos^3(x)(x+1)}- \frac{1}{x^2+1}$
and take the limit of each fraction separately.