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)³ - 1³) / (x^1/2 - 1)

 =  lim _x->1 √x⋅ [lim _x->1 ((x)^3/2 - 1^3/2)]/[lim _x->1(x^1/2 - 1^1/2)]

 =  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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