# CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Ex 14(a)

Odisha State Board Elements of Mathematics Class 11 Solutions CHSE Odisha Chapter 14 Limit and Differentiation Ex 14(a) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 14 Limit and Differentiation Exercise 14(a)

Question 1.
$$\lim _{x \rightarrow 3}$$(x + 4)
Solution:
Clearly, if we take x very close to 3, x + 4 will go very close to 7.
Now let us use ε – δ technique to confirm the result.
Given ε > 0, we seek for δ > 0 depending on ε such that
|x – 3| < δ ⇒ |(x + 4) – 7|< ε
Now |(x + 4) – 7| < ε
if |x – 3| < ε
∴ We can choose ε = 8
Hence for given ε > 0, there exist 8 = ε > 0
such that |x – 3| < δ ⇒ |(x + 4) – 7| < ε
∴ $$\lim _{x \rightarrow 3}$$(x + 4) = 7

Question 2.
$$\lim _{x \rightarrow 1}$$(4x – 1)
Solution:
By taking very close to 1 we have 4x- 1 tends to 3.
Let us use ε – δ technique to confirm the result.
Given ε > 0. We shall find δ > 0 depending on ε such that
|x – 1| < 5 ⇒ |(4x – 1) – 3| < ε
Now |(4x – 1 ) – 3| < ε
if |4x – 1| < ε i.e.|x – 1| < $$\frac{\varepsilon}{4}$$
Let us choose δ = $$\frac{\varepsilon}{4}$$
∴ For given ε > 0 there exists δ = $$\frac{\varepsilon}{4}$$ > 0
such that |x – 1| < δ
⇒ |(4x – 1) – 3| < ε
∴ $$\lim _{x \rightarrow 1}$$(4x – 1) = 3

Question 3.
$$\lim _{x \rightarrow 1}$$(√x + 3)
Solution:
As x → 1 we see √x + 3 → 4
We will confirm the result using ε – δ technique
Let ε > 0, we will choose δ > 4
such that |x – 1| < 8 ⇒ |√x + 3 – 4| < ε
Now |√x + 3 – 4| = |√x – 1|
$$=\frac{|x-1|}{|\sqrt{x}+1|}$$
But |√x + 1| > 1
⇒ $$\frac{1}{|\sqrt{x}+1|}$$ < 1
⇒ $$\frac{|x-1|}{|\sqrt{x}+1|}<\frac{\delta}{1}$$
∴ (√x + 3) – 4 < $$\frac{\delta}{1}$$
We can take δ < ε i.e. δ = min {1, ε}
∴ |x – 1| < δ ⇒ |(√x + 3) – 4| < ε
for given ε > 0 and (δ = ε)
⇒ $$\lim _{x \rightarrow 1}$$(√x + 3) = 4

Question 4.
$$\lim _{x \rightarrow 0}$$ (x2 + 3)
Solution:
As x → 0 we observe that x3 + 3 → 3
Let us use ε – δ technique to confirm the result.
Let ε > 0, we seek for a δ > 0 such that
|x – 0| < ε ⇒ |x2 + 3 – 3| < ε
Let |x| < 8
Now |x2 + 3 – 3| < ε
We have |x|2 < ε ⇒| x| < √ε
(∴ |x| and ε are positive.)
∴ we can choose δ = √ε
∴ We have for given δ > 0 there exists
δ = √ε > 0 such that |x| < δ ⇒ |x2 + 3 – 3| < ε
∴ $$\lim _{x \rightarrow 0}$$ (x2 + 3) = 3

Question 5.
$$\lim _{x \rightarrow 0}$$ 7
Solution:
If x → 0 we observe that 7 → 7.
Let us use e- 8 technique to confirm the limit.
Let f(x) = 7
Given ε > 0, we will choose a δ > 0
such that |x – 0| < δ ⇒ |f(x) – 7| < ε
Now |f(x) – 7| < ε
If f(x) ∈ (7 – ε . 7 + ε)
But for every x, f(x) = 7
⇒ for|x| < δ also f(x) = 7 ∈ (7 – ε . 7 + ε)
∴ Choosing ε = δ we have
|x| < δ ⇒ |f(x) – 7| < ε
∴ $$\lim _{x \rightarrow 0}$$ (7) = 7

Question 6.
$$\lim _{x \rightarrow 1} \frac{(x-1)^3}{(x-1)^3}$$
Solution:
We guess the limit is 1.
Let us confirm using ε – δ technique.
Let ε > 0, f(x) = $$\frac{(x-1)^3}{(x-1)^3}$$
We will choose a δ > 0 such that
|x – 1| < δ ⇒ |f(x) – 1)| < ε
Now |f(x) – 1| < ε
if 1 – ε < f(x) < 1 + ε
∴ We will choose a δ > 0 such that
x ∈ (1 – δ, 1 + δ) – { 1 }
⇒ f(x) ∈ ( 1 – ε, 1 + ε)
As f(x) = for x ≠ 1
We have f(x) ∈ (1 – ε. 1 + ε) for all x ∈ (1 – δ, 1 + δ) – [1]
∴ We can choose δ = ε
for given ε > 0, there exists δ = ε
s.t. |x – 1| < δ ⇒ |f(x) – 1| < ε
∴ $$\lim _{x \rightarrow 1}$$ f(x) = 1

Question 7.
$$\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}$$
Solution:
If we take x very close to 3 (≠ 3)
we have $$\frac{x^3-9}{x-3}$$
= $$\frac{(x-3)\left(x^2+3 x+3^2\right)}{2}$$ → 27
Let ε > 0 and x ≠ 3
Now |$$\frac{x^3-9}{x-3}$$ – 27| = |x2 + 3x +9 – 27|
=|x2 – 9 + 3(x – 3)| ≤ |x2 – 9| + 3|x – 3|
= |x – 3| [|x + 3| + 3] ≤ |x – 3| [|x + 6| < |x – 3| [|x – 3 + 9|]]
If |x – 3| < δ and δ < 1 then |x – 3| [x – 3 + 9| < δ {1 + 9} = 10 δ
Let δ = min {1, $$\frac{\varepsilon}{10}$$}
∴ For given ε > 0 we have a δ = min {1, $$\frac{\varepsilon}{10}$$} >0 such that
|x – 3| < δ ⇒ |$$\frac{x^3-9}{x-3}$$ – 27|
∴ $$\lim _{x \rightarrow 3} \frac{x^3-9}{x-3}$$ = 27

Question 8.
$$\lim _{x \rightarrow 1} \frac{3 x+2}{2 x+3}$$
Solution:
we observe that as x → 1, $$\frac{x+2}{2 x+3}$$ → 1
To establish this let ε > 0,
we seek a δ > 0,

Question 9.
$$\lim _{x \rightarrow 0}|x|$$
Solution:
We see that when x → 0,|x| → 0
Let us establish this using ε – δ technique.
Let ε > 0 we seek a δ > 0 depending on
ε s.t.|x – 0| < ε ⇒ ||x| – 0| < ε
Now ||xl – 0| = ||x|| = |x| < δ
By choosing ε = δ we have |x| < ε ⇒ ||x| – 0| < ε
∴ $$\lim _{x \rightarrow 0}|x|$$ = 0

Question 10.
$$\lim _{x \rightarrow 2}(|x|+3)$$
Solution:
We see that as x → 2, |x| + 3 → 5
Let ε > 0 we were searching for a, δ > 0
such that |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
Now ||x|| + 3 – 5| = ||x| – 2| < |x – 2| < δ
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ Choosing ε = δ
We have |x – 2| < δ ⇒ ||x| + 3 – 5| < ε
∴ $$\lim _{x \rightarrow 2}(|x|+3)$$ = 5