Explain in detail how to find these limits

lim x->π/4 (1-tanx)/(sinx-cosx)

I know you have to change the tan to sin/cos...but this is where I get lost:

"Rewrite tanx =sinx/cosx, then multiply each term by cosx

(Cosx -sinx)
-------------------
Cosx( sinx-cosx) =

-1/cosx

=-secx"
...How do you get (Cosx -sinx)/ (sinx-cosx) ? And where does the 1 go?!

And then there is 2) lim x->0 (4(e^(2x)-1))/(e^x - 1).

Factor :4(e^x-1)(e^x+1)/ (e^x-1)

=4(e^x+1)

Then lim( x->0) = 4(2)=8...

...I don't understand that process at all. Please help a distressed Calculus 1 student understand...

For the first one, when when you multiply (1-tanx) by cosx you get cosx-sinx. Here is why:

cosx(1-tanx) = cosx*1 - cosx*tanx = cosx - cosx*(sinx/cosx)

cosx is canceled in the second term, and you get cosx - sinx.

So when you multiply the numerator by cosx, you get cosx-sinx
And when you multiply denominator by cosx, you get cosx(sinx-cosx).

Now you have (cosx-sinx)/cosx(sinx-cosx)
You can write sinx-cosx as -(cosx-sinx)
(cosx-sinx) is canceled, and you are left with -1/cosx

In the second problem you should use the identity a^2-b^2 = (a-b)(a+b)

e^(2x) is actually (e^x)^2

So e^(2x)-1 = (e^x)^2 - (1)^2 (compare with the identity;here a=e^x and b=1)

(e^x)^2 - (1)^2 = (e^x-1)(e^x+1)

Again, when you divide it by the denominator, e^x-1 is canceled out, and you are left with 4(e^x+1)

Hope that helps!