Evaluate the following limits in Exercises 1 to 22.
Question 1:
Solution:
In
, as x⇢3 Put x = 3, we get
= 3+3 = 6
Question 2:
Solution:
In
, as x⇢π Put x = π, we get
=
Question 3:
Solution:
In
, as r⇢1 Put r = 1, we get
= π
Question 4:
Solution:
In
, as x⇢4 Put x = 4, we get
=
Question 5:
Solution:
In
, as x⇢-1 Put x = -1, we get
=
=
Question 6:
Solution:
In
, as x⇢0 Put x = 0, we get
As, this limit becomes undefined
Now, let’s take x+1=p and x = p-1, to make it equivalent to theorem.
As, x⇢0 ⇒ p⇢1
Here, n=5 and a = 1.
= 5(1)4
= 5
Question 7:
Solution:
In
, as x⇢2 Put x = 2, we get
As, this limit becomes undefined
Now, let’s Factorise the numerator and denominator, we get
=
Cancelling (x-2), we have
=
Put x = 2, we get
=
Question 8:
Solution:
In
, as x⇢3 Put x = 3, we get
As, this limit becomes undefined
Now, let’s Factorise the numerator and denominator, we get
=
=
Cancelling (x-3), we have
=
Put x = 3, we get
=
=
=
Question 9:
Solution:
In
, as x⇢0 Put x = 0, we get
= b
Question 10:
Solution:
In
, as z⇢1 Put z = 1, we get
Let’s take
= p and = p2, As, z⇢1 ⇒ p⇢1
=
Now, let’s Factorise the numerator, we get
=
Cancelling (p-1), we have
=
Put p = 1, we get
= 2
Question 11:
Solution:
In
, as x⇢1 Put x = 1, we get
=
= 1 (As it is given a+b+c≠0)
Question 12:
Solution:
In
, as x⇢-2 Firstly, lets simplify the equation
Cancelling (x+2),we get
Put x = -2, we get
=
Question 13:
Solution:
In
, as x⇢0 Put x = 0, we get
As, this limit becomes undefined
Now, let’s multiply and divide the equation by a, to make it equivalent to theorem.
Hence, we have
=
=
As x⇢0, then ax⇢0
=
By using the theorem, we get
=
=
Question 14:
Solution:
In
, as x⇢0 Put x = 0, we get
As, this limit becomes undefined
Now, let’s multiply and divide the numerator by ax and denominator by bx to make it equivalent to theorem.
Hence, we have
=
=
By using the theorem, we get
=
=
Question 15:
Solution:
In
, as x⇢π Put x = π, we get
As, this limit becomes undefined
Now, let’s take π-x=p
As, x⇢π ⇒ p⇢0
=
=
By using the theorem, we get
=
=
Question 16:
Solution:
In
, as x⇢0 Put x = 0, we get
=