Stuck on derivate/limit problems

[yossarian]

Ok I have two questions, the first one is more curiousity, as it is a problem I couldn't answer on a test, and just want to know how I could solve it as a matter of curiosity. The second one involves the derivate of a trig function.

Problem one:
What is the limit as X goes to 1 of
(1-x^(1/2))/(1-x)
1 minus the square root of x divided by 1 minus x.
I tried every trick I was aware of to try and factor out the x on the bottom, but I had no luck. Any ideas?

Problem two:
what is dy/dx of
y=(sec(x)+tan(x))*(sec(x)-tan(x))

I've tried multiplying it out then working with that, I've tried turning it into (sin(x)+1)/cos(x) and (1-sin(x))/cos(x) and going from there, I've screwed around with the trig identities and the known derivates of the trig functions for a couple of hours, but have not been able to get the correct answer, which is 0. I would really appreciate some help, thanks in advance.

 

[galactus]

Ok I have two questions, the first one is more curiousity, as it is a problem I couldn't answer on a test, and just want to know how I could solve it as a matter of curiosity. The second one involves the derivate of a trig function.

Problem one:
What is the limit as X goes to 1 of
(1-x^(1/2))/(1-x)
1 minus the square root of x divided by 1 minus x.
I tried every trick I was aware of to try and factor out the x on the bottom, but I had no luck. Any ideas?

Multiply top and bottom by the conjugate of the numerator:

\(\displaystyle \L\\\lim_{x\to\1}\frac{1-\sqrt{x}}{1-x}\cdot\frac{1+\sqrt{x}}{1+\sqrt{x}}\)

\(\displaystyle \L\\\lim_{x\to\1}\frac{\sout{1-x}}{(1+\sqrt{x})\sout{(1-x)}}\)

\(\displaystyle \L\\\lim_{x\to\1}\frac{1}{1+\sqrt{x}}\)

Now, see what the limit is?.

Problem two:
what is dy/dx of
y=(sec(x)+tan(x))*(sec(x)-tan(x))

Expand out and see what you have:

\(\displaystyle \L\\sec^{2}(x)-sec(x)tan(x)+sec(x)tan(x)-tan^{2}(x)\)

\(\displaystyle \L\\sec^{2}(x)-tan^{2}(x)=1\)

The derivative of 1 is..................yes............0.

Voila!

 

[yossarian]

Doh! Thanks, I really should have gotten that first problem.... and thanks for the help on the second problem.