# LIMIT OF A FUNCTION EXAMPLES WITH ANSWERS

Limit of a Function Examples With Answers :

Here we are going to see some example questions on evaluating limits.

To find the formulas please visit "Formulas in evaluating limits".

## Calculating limits of a function- Examples

Question 1 :

Evaluate the following limit

lim  x -> 0 √(x2 + a2) - a / √(x2 + b2) - b

Solution :

=  lim  x -> 0 √(x2 + a2) - a / √(x2 + b2) - b

We may distribute the limits for the numerator and denominator.

=  lim  x -> 0 √(x2 + a2) - a / lim  x -> 0 √(x2 + b2) - b

=  lim x->0 [x2+a2-a2/√(x2+a2)+a)]/[x2+b2-b2/(x2+b2)+b)]

=  lim x->0 [x2/√(x2+a2)+a)]/[x2/(x2+b2)+b)]

=  lim x->0 [x2/√(x2+a2)+a)] ⋅ [(x2+b2)+b)/x2

=  lim x->0 [√(x2+b2)+b)/√(x2+a2)+a)]

=  2b/2a

=  b/a

Hence the value of lim  x -> 0 √(x2 + a2) - a / √(x2 + b2) - b is b/a.

Question 2 :

Evaluate the following limit

lim x-> 0  (2 arc sinx/3x)

Solution :

=  lim x-> 0  (2 sin-1x/3x)

=  (2/3) lim x-> 0  (sin-1x/x)

=  2/3

Hence the value of lim x-> 0  (2 arc sinx/3x) is 2/3.

Question 3 :

Evaluate the following limit

lim x-> 0  (1 – cos x)/x2

Solution :

=  lim x-> 0  (2 sin2(x/2)/(x2(4/4))

=  lim x-> 0  (2/4) sin2(x/2)/(x/2)2

=  lim x-> 0  (2/4) (sin(x/2)/(x/2))2

=  (1/2) lim x-> 0  (sin(x/2)/(x/2))2

=  1/2

Hence the value of lim x-> 0  (1 – cos x)/x2 is 1/2.

Question 4 :

Evaluate the following limit

lim x-> 0  (tan 2x/x)

Solution :

=  lim x-> 0  (tan 2x/x(2/2))

=  lim x-> 0  2 (tan 2x/2x)

=  2 lim x-> 0  (tan 2x/2x)

=  2

Hence the value of  lim x-> 0  (tan 2x/x) is 2.

Question 5 :

Evaluate the following limit

lim x-> 0  (2x – 3x)/x

Solution :

=  lim x-> 0  (2x – 1 + 1 – 3x)/x

=  lim x-> 0  [(2x – 1) – (3x - 1)]/x

=  [lim x-> 0  (2x – 1)/x] – [lim x-> 0 (3x - 1)/x]

=  log 2 – log 3

=  log (2/3)

Question 6 :

Evaluate the following limit

lim x-> 0  (3x – 1)/√(x+1) – 1

Solution :

=  lim x-> 0  ((3x – 1)/√(x+1) – 1) (√(x+1) + 1 / √(x+1) + 1)

=  lim x-> 0  ((3x – 1)/((x+1) – 1) (√(x+1) + 1)

=  lim x-> 0  ((3x – 1)/x) (√(x+1) + 1)

=  lim x-> 0  ((3x – 1)/x) lim x-> 0 (√(x+1) + 1)

=   log 3 (2)

=  2 log 3

=  log 32

=  log 9

Hence the value of lim x-> 0  (3x – 1)/√(x+1) – 1 is log 9.

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