[SPLIT] limits: cos[(pi/2)+x]/x; sin(x/2)^2/x^2

[rragas]

yea.. i see now.. thanks a lot for helping me.

i have two more questions on this topic.. so i will just post them

lim x=>0 cos[(pi/2)+x]/x

for this one i figured i might need to use a co-function identity, but i didnt know how to manipulate the (pi/2)+x into x-(pi/2).

and

lim x=>0 sin(x/2)^2/x^2

i started by dividing the numerator and denominator by x, but i don't know how to get rid of that square on the sine

 

[pka]

Why not just note that:
\(\displaystyle \L
\cos \left( {x + \frac{\pi }{2}} \right) = - \sin (x)?\)

 

[rragas]

oh yes! i forgot that.

thanks.

And the last one?

 

[stapel]

And the last one?

Try using the trig limits, provided earlier.

Then please reply showing what you have done. Thank you.

Eliz.

 

[rragas]

Ooh. I think I see

So i made it

[sin(x/2)/x]^2

Then i multiply by 1/2 to get (1/2)^2

Right?

Thanks guys. You guys are truely helpful.