How to evaluate each limit below. (a) lim x+2/x^2+x+1 x→∞ (b) lim (1/√x−2 − 4/x−4)? x->4

2 Answers

a)

#\lim_{x\to \infty}\frac{x+2}{x^2+x+1}#

#=\lim_{x\to \infty}\frac{x/x^2+2/x^2}{x^2/x^2+x/x^2+1/x^2}#

#=\lim_{x\to \infty}\frac{1/x+2/x^2}{1+1/x+1/x^2}#

#=\frac{0+0}{1+0+0}#

#=0#

b)

#lim_{x\to 4}\frac{1}{\sqrtx-2}-4/{x-4}#

#=lim_{x\to 4}\frac{x-4-4(\sqrtx-2)}{(\sqrtx-2)(x-4)}#

#=lim_{x\to 4}\frac{x-4-4\sqrtx+8}{(\sqrtx-2)(x-4)}#

#=lim_{x\to 4}\frac{x-4\sqrtx+4}{(\sqrtx-2)(x-4)}#

#=lim_{x\to 4}\frac{(\sqrtx-2)^2}{(\sqrtx-2)(x-4)}#

#=lim_{x\to 4}\frac{\sqrtx-2}{x-4}#

Applying L'Hopital's rule for #0/0# form,

#=lim_{x\to 4}\frac{d/dx(\sqrtx-2)}{d/dx(x-4)}#

#=lim_{x\to 4}\frac{1/{2\sqrtx}}{1}#

#=1/2 lim_{x\to 4}1/{\sqrtx}#

#=1/2 \cdot1/{2}#

#=1/4#

Jul 17, 2018

# 1/4#.

Explanation:

Part (b) can be solved without using L'Hospital's Rule :

Observe that, #x-4=(sqrtx)^2-2^2=(sqrtx+2)(sqrtx-2)#.

#"The Reqd. Lim."=lim_(x to 4){1/(sqrtx-2)-4/(x-4)}#,

#=lim_(x to 4)[1/(sqrtx-2)-4/{(sqrtx+2)(sqrtx-2)}]#,

#=lim_(x to 4)[{(sqrtx+2)-4}/{(sqrtx+2)(sqrtx-2)}]#,

#=lim_(x to 4)(sqrtx-2)/{(sqrtx+2)(sqrtx-2)}#,

#=lim_(x to 4)1/(sqrtx+2)#,

#=1/(sqrt4+2)#,

#=1/4#, as Respected Harish Chandra Rajpoot has readily

derived!