# I don't understand this limit problem

• January 2nd 2010, 02:26 PM
Jones
I don't understand this limit problem
Hi,

I have
$\lim_{x \to 0+} ~\frac{sin^2x}{tan~x-x}$

Now replace the trig functions with its corresponding taylor polynomial

$\frac{\bigg(x-\frac{x^3}{3!}\bigg)^2}{\bigg(x+\frac{x^3}{3}\bigg )-x}$

This limit evaluates to infinity, i don't understand why. It looks more like -2 to me
• January 2nd 2010, 03:15 PM
ANDS!
Too much work (unless you are required to use Taylor representations), both functions approach zero, so you can easily just use L'hopitals rule.
• January 2nd 2010, 07:02 PM
tonio
Quote:

Originally Posted by Jones
Hi,

I have
$\lim_{x \to 0+} ~\frac{sin^2x}{tan~x-x}$

Now replace the trig functions with its corresponding taylor polynomial

$\frac{\bigg(x-\frac{x^3}{3!}\bigg)^2}{\bigg(x+\frac{x^3}{3}\bigg )-x}$

This limit evaluates to infinity, i don't understand why. It looks more like -2 to me

$\frac{\bigg(x-\frac{x^3}{3!}\bigg)^2}{\bigg(x+\frac{x^3}{3}\bigg )-x}=$ $\frac{x^2-\frac{x^4}{6}+\frac{x^6}{36}}{\frac{x^3}{3}}=\frac {3}{x}-\frac{x}{2}+\frac{x^3}{12}\xrightarrow [x\to 0^+]{}\infty$

Tonio
• January 2nd 2010, 10:11 PM
matheagle
or you can use L'Hopital's Rule here, which I just noticed was also recommended.
• January 2nd 2010, 10:21 PM
Drexel28
Quote:

Originally Posted by Jones
Hi,

I have
$\lim_{x \to 0+} ~\frac{sin^2x}{tan~x-x}$

Now replace the trig functions with its corresponding taylor polynomial

$\frac{\bigg(x-\frac{x^3}{3!}\bigg)^2}{\bigg(x+\frac{x^3}{3}\bigg )-x}$

This limit evaluates to infinity, i don't understand why. It looks more like -2 to me

Another convoluted way to show the limit doesn't exist is to assume it does. Then $\lim_{x\to0}\frac{\sin^2(x)}{\tan(x)-x}=L$. But $\lim_{x\to0}\frac{\sin^2(x)}{\tan(x)-x}=\lim_{x\to0}\left\{\frac{1-\cos(x)}{\tan(x)-x}\cdot\left(1+\cos(x)\right)\right\}$. And since $\lim_{x\to0}\left\{1+\cos(x)\right\}=2$, we may by assumption conclude that $\frac{L}{2}=\lim_{x\to0}\frac{1}{1+\cos(x)}\cdot\l im_{x\to0}\left\{\frac{1-\cos(x)}{\tan(x)-x}\cdot\left(1+\cos(x)\right)\right\}$ $=\lim_{x\to0}\left\{\frac{1-\cos(x)}{\tan(x)-x}\cdot\frac{1+\cos(x)}{1+\cos(x)}\right\}=\lim_{x \to0}\frac{1-\cos(x)}{\tan(x)-x}$ which tells us that $\lim_{x\to0}\frac{1-\cos(x)}{\tan(x)-x}$ exists, but this is ludicrous since $1-\cos(x)\sim \frac{x^2}{2}$ and $\tan(x)-x\sim\frac{-x^3}{3}$. Thus, our assumption was false and the original limit does not exist.