Lim (x^3-8)/(sin(pi*x)) x-> 2 Help me please ???

without using hospital method

2 Answers
Dec 11, 2017

See below.

Explanation:

Use:

#x^3-8 = (x-2)(x^2+2x+4)#

and

#sin(pix) = sin(pix-2pi) = sin(pi(x-2))#

#(x^3-8)/sin(pix) = ((x-2)(x^2+2x+4))/sin(pi(x-2))#

# = 1/pi (pi(x-2))/sin(pi(x-2)) (x^2+2x+4)#

#lim_(xrarr2) (x^3-8)/sin(pix) = 1/pi [lim_(xrarr2)(pi(x-2))/sin(pi(x-2))] [lim_(xrarr2)(x^2+2x+4)]#

# = 1/pi * (1) * (12) = 12/pi#

Dec 11, 2017

# 12/pi#.

Explanation:

Let us substitute #x=2+h. :." As "x to 2, h to 0.#

#:." The Reqd. Lim. L="lim_(h to 0){(2+h)^3-8}/sin(pi(2+h)),#

#=lim{(8+12h+6h^2+h^3)-8}/sin(2pi+pih),#

#=lim(12h+6h^2+h^3)/(sinpih),#

#=lim{h(12+6h+h^2)}/(sinpih),#

#={lim_(2h to 0)1/pi*(pih)/(sin2h)}{lim_(h to 0)(12+6h+h^2)},#

#={1/pi*1}{12+0+0}.#

# rArr L=12/pi.#

Enjoy Maths.!