Limits connected to a right angled triangle

Question

Given a right angled triangle with sides $1,x$ and hypotenuse $y$. Let $\theta$ be the angle contained by side $1$ and hypotenuse. Then evaluate the following limits:

$\lim_{\theta\to\pi/2}\sqrt{y}-\sqrt{x}$

$\lim_{\theta\to\pi/2}y-x$

$\lim_{\theta\to\pi/2}y^2-x^2$

$\lim_{\theta\to\pi/2}y^3-x^3$

I was able to evaluate the third one, using $y=\sec{\theta}$ and $x=\tan{\theta}$.

EDIT:

I also solved the first first and second one. Only second and fourth are is unsolved.

How to evaluate others?

Answer 1

First note that$$y^3-x^3=\dfrac{y^6-x^6}{x^3+y^3}=\dfrac{3y^4-3y^2+1}{x^3+y^3}\ge\dfrac{3y^4-3y^2+1}{2y^3}=1.5y-1.5\dfrac{1}{y}+\dfrac{1}{y^3}$$Also $\cos\theta=\dfrac{1}{y}\to 0^+$ leads to $y\to \infty$ and makes the limit $\infty$ either.

Answer 2

I already accepted the answer, but I solved it in a different manner. I am still adding this so that anyone who stumbles upon this in the future will have an alternative.

$$\lim_{\theta\to\pi/2}\sec^3{\theta}-\tan^3{\theta}$$

$$\lim_{\theta\to\pi/2}(\sec{\theta}-\tan{\theta})(\sec^2{\theta}+\tan^2{\theta}+\sec{\theta}\tan{\theta})$$

$$\lim_{\theta\to\pi/2}(\frac{1-\sin{\theta}}{\cos{\theta}})(\frac{1+\sin^2{\theta}+\sin{\theta}}{\cos^2{\theta}})$$

$$\lim_{\theta\to\pi/2}(\frac{1-\sin{\theta}}{\cos{\theta}})(\frac{1+\sin^2{\theta}+\sin{\theta}}{(1-\sin{\theta})(1+\sin{\theta})})$$

$$\lim_{\theta\to\pi/2}(\frac{1+\sin^2{\theta}+\sin{\theta}}{({\cos{\theta}})(1+\sin{\theta})})$$

which gives us $\infty$