Limit using l'Hopitals

[Jaxxo]

I need to calculate this limit: lim(x->0+) [sin(x)]^2 / [tan(x)-x]

This is in the calculus seventh edition, exercise 4.3.17. In this chapter l'hoptial's rules are introduced.

I've derived once: lim(x->0+) [2sin(x)cos(x)] / [1/[cos(x)^2 - 1]] still no limit in sight.

Second time: lim(x->0+) (cos(x)^2 - sin(x)^2) / [sin(x)/cos(x)^3] - possible numerator cos(2x)

I don't see the limit here. It says in the book that from this step they find the limit to be inf+

Please help,
Jacob

 

[HallsofIvy]

I need to calculate this limit: lim(x->0+) [sin(x)]^2 / [tan(x)-x]

This is in the calculus seventh edition, exercise 4.3.17. In this chapter l'hoptial's rules are introduced.

I've derived once: lim(x->0+) [2sin(x)cos(x)] / [1/[cos(x)^2 - 1]] still no limit in sight.

Second time: lim(x->0+) (cos(x)^2 - sin(x)^2) / [sin(x)/cos(x)^3] - possible numerator cos(2x)

This is where you have it. Whether you write the numerator as "cos(2x)" or leave it as "cos^2(x)- sin^2(x)", setting x equal to 0 gives 1.
What happens to the denominator, sin(x)/cos^3(x), when x= 0?

I don't see the limit here. It says in the book that from this step they find the limit to be inf+

Please help,
Jacob

 

[Ishuda]

Just a few things I keep around in my head, for x near zero
sin(x) ~ x + a x³
tan(x) ~ x + b x³
cos(x) ~ 1 + c x²
Sometimes I can even remember what a (=-1/6), b (=1/3), and c (=-1/2} are other than just non-zero. In any case we note that the original limit should result in something like

\(\displaystyle \frac{sin^2(x)}{tan(x)-x} \approx \frac{x^2}{b\space x^3} \approx \frac{1}{b\space x}\)

The limit as x goes to zero is not going to behave nicely. Now look at that first application of L'hoptial's rule. That is we want the limit of
\(\displaystyle \frac{2\space sin(x)\space cos(x)}{\frac{1}{cos^2(x)}\space -\space 1}
= \frac{2\space sin(x)\space cos^3(x)}{1\space -\space cos^2(x)}
= \frac{2\space sin(x)\space cos^3(x)}{sin^2(x)}
= \frac{2\space cos^3(x)}{sin(x)}
\approx\space\frac{2}{x}\)

I think it's time to stop.