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.

\(\lim _{x \rightarrow 0}\) (x^{2} + 3)

Solution:

As x → 0 we observe that x^{3} + 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 ⇒ |x^{2} + 3 – 3| < ε

Let |x| < 8

Now |x^{2} + 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 ⇒ |x^{2} + 3 – 3| < ε

∴ \(\lim _{x \rightarrow 0}\) (x^{2} + 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| = |x^{2} + 3x +9 – 27|

=|x^{2} – 9 + 3(x – 3)| ≤ |x^{2} – 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