How to solve #lim_(x->0)(sqrt(cosx)-1)/x^2# ?

3 Answers
Dec 11, 2017

#-1/4#

Explanation:

#(sqrt(cosx)-1)/x^2 = (cosx-1)/(x^2 (sqrtcosx+1))#

but for small #abs x# we have #cosx approx 1-x^2/(2!)+O(x^4)# then

#lim_(x->0)(sqrt(cosx)-1)/x^2 = lim_(x->0) (-x^2/(2!)+O(x^4))/(x^2(sqrt cosx+1)) = -1/4#

Dec 11, 2017

# -1/4#.

Explanation:

We assume that,

#x in [-pi/2,pi/2]," so that "cosx ge 0," making "sqrtcosx" meaningful."#

#"The Reqd. Lim. L="lim_(x to 0){sqrtcosx-1)/x^2#,

#=lim_(x to 0){sqrtcosx-1)/x^2xx(sqrtcosx+1)/(sqrtcosx+1)#,

#=lim_(x to 0)-(1-cosx)/{x^2(sqrtcosx+1)#,

#=-lim_(x to 0)(2sin^2(x/2))/{x^2(sqrtcosx+1)#,

#=-lim_(x to 0){2/(sqrtcosx+1)}{1/x^2(sin(x/2))^2/(x/2)^2*(x/2)^2},#

#=-lim_(x to 0){2/(sqrtcosx+1)}{sin(x/2)/(x/2)}^2*1/4,#

#=-2/(sqrtcos0+1)*(1)^2*1/4,#

#=-2/(1+1)*1/4#.

# rArr L=-1/4.#

Dec 11, 2017

#((sqrt(cosx)-1))/x^2 * ((sqrt(cosx)+1))/((sqrt(cosx)+1)) = (cosx-1)/(x^2(sqrt(cosx)+1))#

# = ((cosx-1))/(x^2(sqrt(cosx)+1)) * ((cosx+1))/((cosx +1))#

# = (cos^2x-1)/(x^2(sqrt(cosx)+1)(cosx +1)))#

# = (-sin^2x)/x^2 * 1/((sqrt(cosx)+1)(cosx +1))#

# = -((sinx)/x)^2 * 1/((sqrt(cosx)+1)(cosx +1))#

Evaluating limit as #xrarr0#, we get

# -(1)^1 * 1/((sqrt1+1)(1+1)) = -1/4#