Limits Examples and Solutions :
Here we are going to see some example questions on evaluating limits.
To find the formulas please visit "Formulas in evaluating limits".
Question 1 :
Evaluate the following limit
lim x-> 0 (1 – cos2x)/(x sin 2x)
Solution :
= lim x-> 0 (sin2x)/(2x sin x cos x)
= lim x-> 0 (1/2)(sinx)/(x cos x)
= (1/2) lim x-> 0 (tanx/x)
= (1/2) (1)
= 1/2
Question 2 :
Evaluate the following limit
lim x-> ∞ x (31/x + 1 – cos (1/x) – e1/x)
Solution :
= lim x-> ∞ x (31/x + 1 – cos (1/x) – e1/x)
= lim x-> ∞ (31/x + 1 – cos (1/x) – e1/x)/(1/x)
let y = 1/x
If x -> ∞, then y -> 0
= lim y-> 0 (3y + 1 – cos y – ey)/y
= lim y-> 0 (3y + 1 – cos y – ey + 1 - 1)/y
= lim y-> 0 [(3y - 1) + (1 – cos y) - (ey - 1)]/y
= [lim y-> 0 (3y-1)/y]+[lim y-> 0 (1–cos y)/y]-[lim y-> 0 (ey-1)]/y]
= log 3 + 0 - 1
= log 3 - 1
Hence the value of lim x-> ∞ x (31/x + 1 – cos (1/x) – e1/x) is log 3 - 1.
Question 3 :
Evaluate the following limit
lim x-> ∞ {x [log (x + a) – log x] }
Solution :
= lim x-> ∞ {x [log (x + a) – log x] }
= lim x-> ∞ {x [log (x + a)/x] }
= lim x-> ∞ (log (1 + a/x))/(1/x)
= lim x-> ∞ (a/a)(log (1 + a/x))/(1/x)
= lim x-> ∞ a(log (1 + a/x))/(a/x)
Let y = a/x
If x -> ∞, then y -> 0
= lim y -> 0 a(log (1 + y )/y)
= a
Question 4 :
Evaluate the following limit
lim x-> π sin 3x/sin 2x
Solution :
= lim x-> π sin 3x/sin 2x
By applying the limit in the given question, we get 0/0.
= lim x-> π sin (2x + x)/sin 2x
= lim x-> π (sin 2x cos x + cos 2x sin x)/sin 2x
= lim x-> π [(sin 2x cos x)/sin 2x + (cos 2x sin x)/sin 2x]
Applying the formula for sin 2x = 2 sin x cos x
= lim x-> π cos x + lim x-> π (cos 2x sin x)/(2 sin x cos x)
= cos π + cos 2π/2cos π
= -1 - (1/2)
= -3/2
Hence the value of lim x-> π sin 3x/sin 2x is -3/2.
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