# EVALUATING LIMITS WITH SQUARE ROOTS

## About "Evaluating Limits With Square Roots"

Evaluating Limits With Square Roots :

Here we are going to see some practice problems on evaluating limits with solutions.

## Evaluating Limits With Square Roots - Questions

Question 1 :

Evaluate

lim x->2 (1/x) - (1/2) / (x - 2)

Solution :

lim x->2 (1/x) - (1/2) / (x - 2)

=  lim x->2 (2 - x)/2x(x - 2)

=  lim x->2 -(x - 2)/2x(x - 2)

=  lim x->2 -(1/2x)

By applying the value of x, we get

=  -(1/2(2))

=   -1/4

Hence the value of lim x->2 (1/x) - (1/2) / (x - 2) is -1/4.

Question 2 :

Evaluate

lim x->1 (√x - x2) / (1 - √x)

Solution :

lim x->1 (√x - x2) / (1 - √x)

=  lim x->1 (√x - (x)4) / (1 - √x)

=  lim x->1 √x (1 - (√x)3) / (1 - √x)

=  lim x->1 √x (13 - (√x)3) / (1 - √x)

=  lim x->1 √x  ⋅ lim x->1 (13 - (√x)3) / (1 - √x)

=  lim x->1 √x  ⋅ lim x->1 ((√x)3 - 13) / (√x - 1)

=  lim x->1 √x  ⋅ lim x->1 ((√x)- 13) / (x1/2 - 1)

=  lim x->1 √x⋅ [lim x->1 ((x)3/2 - 13/2)]/[lim x->1(x1/2 - 11/2)]

=  lim x->1 √x (3/2) / (1/2)

=  1 ⋅ (3/2) ⋅ (1/2)

=  3

Hence the value of lim x->1 (√x - x2) / (1 - √x) is 3.

Question 3 :

Evaluate

lim x->0 (√(x2 + 1) - 1) / (√(x2 + 16) - 4)

Solution :

lim x->0 (√(x2 + 1) - 1) / (√(x2 + 16) - 4)

=  [lim x->0 (√(x2 + 1) - 1)] / [lim x-> 0 (√(x2 + 16) - 4)] By applying the limit values, we get

=  (√16 + 4)/(√1 + 1)

=  (4 + 4)/(1 + 1)

=  8/2

=  4

Hence the value of lim x->0 (√(x2 + 1) - 1) / (√(x2 + 16) - 4) is 4.

Question 4 :

lim x->0 (√(1 + x) - 1)/x

Solution :

lim x->0 (√(1 + x) - 1)/x

=  lim x->0 [(√(1 + x) - 1)/x]⋅ [(√(1 + x) + 1)/(√(1 + x) + 1)]

=  lim x->0 [(√(1 + x) - 1)/x]⋅ [(√(1 + x) + 1)/(√(1 + x) + 1)]

=  lim x->0 [(√(1 + x)2 - 12)/x(√(1 + x) + 1)]

=  lim x->0 [ x2/x(√(1 + x) + 1)]

=  lim x->0 [ x/(√(1 + x) + 1)]

By applying the limit value, we get

=  0/√1 + 1

=  0

Hence the value lim x->0 (√(1 + x) - 1)/x is 0.

Question 5 :

lim x->1 ((7 + x3) - (3 + x2))/(x - 1)

Solution :

lim x->1 ((7 + x3) - (3 + x2))/(x - 1)

=  lim x->1 ((7 + x3) - (3 + x2))/(x - 1)

=  lim x->1 ((7 + x3) - 2 + 2 - (3 + x2))/(x - 1)

=  lim x->1 [(7+x3)-2)/(x-1)]-lim x->1[(√(3+x2)-2)/(x-1)] =  lim x->1(x3-1)(1/3)8(-2/3)/(x-1) - lim x->1(x2-1)/(x-1)(1/2) 4(-1/2)

=  lim x->1(x2 + x + 1)(1/3)(1/4) - lim x->1(x+1)(1/2) (1/2)

=  1/4 - 1/2

=  - 1/4

Hence the value of lim x->1 ((7 + x3) - (3 + x2))/(x - 1) is 1/4. After having gone through the stuff given above, we hope that the students would have understood, "Evaluating Limits With Square Roots"

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