Find limit Lim [√ (2x + 1) - 3] / √ X - 2, X - > 4

Find limit Lim [√ (2x + 1) - 3] / √ X - 2, X - > 4

The answer is 4 / 3
Because both the numerator and denominator have factors that cause the fraction to become zero, the numerator and denominator should be rationalized separately to eliminate the root sign
lim[x→4] [√(2x+1)-3]/(√x-2)
=Lim [x → 4] {[√ (2x + 1) - 3] [√ (2x + 1) + 3] (√ x + 2)} / {(√ X-2) (√ x + 2) [√ (2x + 1) + 3]}. There are three terms in the molecule and three terms in the denominator. Here, multiply one term and divide by another term. Don't forget to multiply without dividing
=lim[x→4] [(2x+1-9)(√x+2)]/{(x-4)[√(2x+1)+3]}
=lim[x→4] [2(x-4)(√x+2)]/{(x-4)[√(2x+1)+3]}
=2lim[x→4] (√x+2)/[√(2x+1)+3]
=2·(√4+2)/[√(2·4+1)+3]
=2·4/6
=4/3

Find LIM (x ^ 3 + 2x) / (x-1) ^ 2, the limit when x tends to 1

It's infinite
When x → 1,
Molecule X & # 179; + 2x = 3
And the denominator (x-1) &# 178; = 0
3/0=∞

The limit of LIM (x → 1) (x-x ^ x) / (1-x + INX)?

lim(x→1）(x-x^x)/(1-x+Inx)=lim(x→1）x(1-x^(x-1)/(1-x+lnx)=lim(x→1）(1-x^(x-1))/(1-x+lnx)=lim(x→1）(x^(x-1)(lnx+1-1/x)/(-1+1/x)=lim(x→1）(xlnx+x-1)/(-x+1)=lim(x→1）(lnx+1+1)/-1=-2