Question #ba3fd

3 Answers
Cesareo R.
Jun 23, 2017

#1/2#

Explanation:

Defining

#f(y) = ((y - 1)^(3/2) - 1)/sqrt[y]# we have

#(f(1+x) - f(1))/x=((x sqrt[x] - 1)/sqrt[x + 1] + 1)/x# and then

#lim_(x->0)((f(1+x) - f(1))/x)=f'(1) = 1/2#

NOTE:

#d/(dx)f(x) = ((2x+1)sqrt(x-1)+1)/(2x^(3/2)# so

#f'(1) = 1/2#

Ratnaker Mehta
Jun 24, 2017

# 1/2.#

Explanation:

The Reqd. Lim. =#lim_(x to 0){(xsqrtx-1)/sqrt(x+1)+1}/x,#

#=lim_(x to 0) 1/x{(xsqrtx-1)/sqrt(x+1)+1},#

#=lim_(x to 0) 1/x{(xsqrtx)/sqrt(x+1)-1/sqrt(x+1)+1},#

#=lim_(x to 0) (cancelxsqrtx)/(cancelxsqrt(x+1))+1/x{1-1/sqrt(x+1)},#

#=lim_(x to 0) sqrtx/sqrt(x+1)+1/x{(sqrt(x+1)-1)/sqrt(x+1)},#

#=sqrt0/sqrt(0+1)+lim_(x to 0)1/x{(sqrt(x+1)-1)/sqrt(x+1)xx(sqrt(x+1)+1)/(sqrt(x+1)+1)},#

#=lim_(x to 0)1/x{((x+1)-1)/((x+1)+sqrt(x+1))},#

#=lim_(x to 0) cancelx/{cancelx((x+1)+sqrt(x+1))},#

#=lim_(x to 0) 1/((x+1)+sqrt(x+1)),#

#=1/((0+1)+sqrt(0+1))=1/(1+1),#

# rArr" The Reqd. Lim.="1/2,# as, Respected Cesareo R. Sir has

readily obtained.

Enjoy Maths.!

Ratnaker Mehta
Jun 24, 2017

# 1/2.#

Explanation:

There is still another Method to solve the Problem.

We subst. #x=tan^2y rArr" As "x to 0, y to 0.#

#:." The Reqd. Lim.="lim_(x to 0) ((xsqrtx-1)/sqrt(x+1)+1)/x,#

#=lim_(y to 0){(tan^2ysqrt(tan^2y)-1)/sqrt(tan^2y+1)+1}/tan^2y,#

#=lim_(y to 0)((tan^3y-1)/secy+1)/tan^2y,#

#=lim_(y to 0){(sin^3y-cos^3y)/(cos^3y*secy)+1}/(sin^2y/cos^2y),#

#=lim_(y to 0){(sin^3y-cos^3y)/cos^2y+1}(cos^2y/sin^2y),#

#=lim_(y to 0)(sin^3y-cos^3y+cos^2y)/cos^2yxxcos^2y/sin^2y,#

#=lim_(y to 0) (sin^3y-cos^3y+cos^2y)/sin^2y,#

#=lim_(y to 0) sin^3y/sin^2y+(cos^2y-cos^3y)/sin^2y,#

#=lim_(y to 0) siny+cos^2y/sin^2y*(1-cosy),#

#=lim_(y to 0) siny+cos^2y/sin^2y*((1-cosy)(1+cosy))/(1+cosy),#

#=lim_(y to 0) siny+cos^2y/sin^2y*sin^2y/(1+cosy),#

#=lim_(y to 0) siny+cos^2y/(1+cosy),#

#=sin0+cos^2 0/(1+cos0),#

#=0+1^2/(1+1).#

# rArr" The Reqd. Lim.="1/2,# as before!

Enjoy Maths.!

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