Posted by **Alex** on Saturday, February 21, 2009 at 6:39pm.

I have a question relating to limits that I solved

lim(x-->0) (1-cosx)/2x^2

I multiplied the numerator and denominator by (1+cosx) to get

lim(x->0) (1-cos^2x)/2x^2(1+cosx)

= lim(x->0)sin^2x/2x^2(1+cosx)

the lim(x->0) (sinx/x)^2 would =1

I then substituted the remaining portion with 0 and got 1/4

Is this answer right??

## Answer This Question

## Related Questions

- Calculus - Find the following limits algebraically or explain why they don’t ...
- maths - 1)lim [(x+y)sec(x+y)-xsecx]/ y x-o 2)show that lim |x-4|/x-4 does not ...
- Trigonometry. - ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! ...
- Trig........ - I need to prove that the following is true. Thanks (cosx / 1-sinx...
- Pre-Calc - Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= ...
- Trigonometry Check - Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [...
- calculus - using L'Hospital's rule, evalutate; LIM as x->0 e^x +cos x / e^x...
- Precalculus/Trig - I can't seem to prove these trig identities and would really ...
- Calculus - How do I solve lim(x-->pi/4)(sinx-cosx)/cos(2x) The (x-->pi/4) ...
- CALCULUS - Could someone please solve these four problems with explanations? I'd...

More Related Questions