Evaluating Limits With Square Roots :
Here we are going to see some practice problems on evaluating limits with solutions.
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) / (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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