I've got 2, and I'm not sure of the correct way to type them, so here it goes.

lim as x->Pi/4 (1-tan(x))/(sin(x)-cos(x))

lim as x->0 (sin(x^2))/(sin^2(x))

any help is greatly appreciated.

Are you familiar with L'Hopital's rule?

I am, but I'm in a class where we haven't even gotten to derivatives yet.
I've had a previous course, so I know how, but for now I'm stuck on solving these the algebraic way.

The first poblem can be solved if you replace the denominator as follows:

sin(x) - cos(x) = cos(x)(sin(x)/cos(x)-1) = cos(x)(tan(x) - 1).

Can you take it from there?

Still thinking about the 2nd one...

For the 2nd one, you can use an infinite series substitution. I realize this is not an algebraic solution, and may be beyond the level you're at, but here goes:

Given sin(a) = a - a^3/3! + a^5/5! -...
Then sin(a^2) = a^2 - a^6/3! + a^10/5! -...
and sin^2(a) = (a - a^3/3! + a^5/5! -... )*(a - a^3/3! + a^5/5! -... )
= a^2 - 2a^4/3! + ^6/(3!*3!) -...

As lim(a) ->0, throw out the higher ordered terms (since they go to zero much faster than the lower order terms) and you get

lim as a->0 = a^2/a^2 = 1.

The first problem whittles down to

\frac{1-tan(x)}{sin(x)-cos(x)}=\frac{-1}{cos(x)}

Therefore, you have \lim_{x\to\frac{\pi}{4}}\frac{-1}{cos(x)}

Now, it's easy. Right?

For the 2nd problem, heres' a different approach. You may have already covered in class the fact that:

\lim_ {x \to 0} \frac {\sin(x)} x = 1.

If you have that, then it follows that:

(a)\ \ \lim_ {x \to 0} \frac {\sin(x^2)} {x^2} = 1.

Also,

(b)\ \ \lim_ {x \to 0} \( \frac {\sin (x)} {x} \) ^2= 1.


Divide (a) by (b), and you've got it.