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

Question

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

Isaiah Allen · Accepted Answer

a)  #\lim_{x	o \infty}\frac{x+2}{x^2+x+1}# #=\lim_{x	o \infty}\frac{x/x^2+2/x^2}{x^2/x^2+x/x^2+1/x^2}# #=\lim_{x	o \infty}\frac{1/x+2/x^2}{1+1/x+1/x^2}# #=\frac{0+0}{1+0+0}# #=0# b) #lim_{x	o 4}\frac{1}{\sqrtx-2}-4/{x-4}# #=lim_{x	o 4}\frac{x-4-4(\sqrtx-2)}{(\sqrtx-2)(x-4)}# #=lim_{x	o 4}\frac{x-4-4\sqrtx+8}{(\sqrtx-2)(x-4)}# #=lim_{x	o 4}\frac{x-4\sqrtx+4}{(\sqrtx-2)(x-4)}# #=lim_{x	o 4}\frac{(\sqrtx-2)^2}{(\sqrtx-2)(x-4)}# #=lim_{x	o 4}\frac{\sqrtx-2}{x-4}# Applying L'Hopital's rule for #0/0# form, #=lim_{x	o 4}\frac{d/dx(\sqrtx-2)}{d/dx(x-4)}# #=lim_{x	o 4}\frac{1/{2\sqrtx}}{1}# #=1/2 lim_{x	o 4}1/{\sqrtx}# #=1/2 \cdot1/{2}# #=1/4#

Evelyn Agard · Answer

# 1/4#.

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!

Axel Alvarez · Answer

undefined (a) To evaluate $ \lim_{x \to \infty} \frac{x+2}{x^2+x+1} $, divide both the numerator and denominator by the highest power of $ x $, which is $ x^2 $. This yields $ \lim_{x \to \infty} \frac{\frac{x}{x^2}+\frac{2}{x^2}}{1+\frac{x}{x^2}+\frac{1}{x^2}} $. As $ x $ approaches infinity, all terms with $ x $ in the denominator approach zero. Thus, the limit simplifies to $ \frac{0+0}{1+0+0} = \frac{0}{1} = 0 $.

(b) To evaluate $ \lim_{x \to 4} \left(\frac{1}{\sqrt{x}-2} - \frac{4}{x-4}\right) $, first note that both expressions are undefined at $ x = 4 $ due to division by zero. However, we can find the limit using algebraic manipulation. Rationalize the first expression by multiplying the numerator and denominator by the conjugate of the denominator. This gives $ \lim_{x \to 4} \left(\frac{1}{\sqrt{x}-2} \times \frac{\sqrt{x}+2}{\sqrt{x}+2} - \frac{4}{x-4}\right) $. After simplification, the expression becomes $ \lim_{x \to 4} \frac{(\sqrt{x}+2)-(4(\sqrt{x}+2))}{x-4} $. Further simplification yields $ \lim_{x \to 4} \frac{-3\sqrt{x}}{x-4} $. Substituting $ x = 4 $ gives $ \frac{-3\sqrt{4}}{4-4} = \frac{-6}{0} $, which is undefined. Therefore, the limit does not exist.

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

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