Samacheer Class 11 Maths - Limits and Continuity

\(\lim _{x \rightarrow 1}\) sin πx
From the graph x = 1, the curve y = f(x) intersects the line x = 1 at x – axis.
∴ y = f(1) = 0
Hence \(\lim _{x \rightarrow 1}\) sin πx = 0

To find \(\lim _{x \rightarrow 0}\) sec x
Let y = f(x) = sec x
From the graph at x = 0 the curve intersect the y – axis.
At x = 0 we have y = 1
∴ \(\lim _{x \rightarrow 0}\) sec x = 1

No, f(x) = 4, It is the value of the function at x = 2
This limit doesn’t exists at x = 2
Since f(2) = 4
It need not imply that \(\lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x)\)
∴ we cannot conclude at x = 2

\(\lim _{x \rightarrow 2}\) f(x) 4, \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x) = 4
When x approaches 2 from the left or from the right f(x) approaches 4.
Given that \(\lim _{x \rightarrow 2^{-}}\) f(x) = \(\lim _{x \rightarrow 2^{+}}\) f(x) = 4
The existence or non-existence at x =2 has no leaving on the existence of the limit of f(x) as x approaches to 2.
∴ We cannot conclude the value of f(2).

Given R(s) = \(\frac{\mathbf{s}}{(\alpha \mathbf{S}+\beta)}\)
where S is the number of spawners, R is the number of recruits. α, β are positive constants.
When the number of spawners is sufficiently large s → ∞
When the number of spawners is sufficiently large, the number of recruits is \(\frac{1}{\alpha}\)

f(x) = 2x 2 + 3x – 5
Clearly f(x) is defined for all points of R.
Let x 0 be an arbitrary point in R. Then
f(x 0 ) = 2x 0 2 + 3x 0 – 5 ——- (1)
Thus, f(x) is defined at all points of R limit of f(x) exist at all points of R and is equal to the value of the function f (x).
Thus f (x) is continuous at all points of R.

Let f(x) = sin x
f (x) is defined at all points of R.
Let x 0 be an arbitrary point in R.
= x o + sin x 0 …….. (1)
f (x o ) = x o + sin x o ……… (2)
From equations (1) and (2) we get
∴ At all points of R, the limit of f (x) exists and is equal to the value of the function.
Thus, f( x) satisfies ail conditions for continuity.
Therefore, f(x) is continuous at all points of f(x).
(ii) X 2 cos x
Let f(x) = x 2 cos x
f (x) is defined at all points of R.
Let x 0 be an arbitrary point in R. Then
= x 2 0 cos 0
f(x 0 ) = x 2 0 cos 0
From equation (1) and (2), we have
∴ The limit at x = x 0 exist and is equal to the value of the function f(x) at x = x 0.
Since x 0 is arbitrary, the limit of the function exist and is equal to the value of the function for all points in R.
∴ f( x) satisfies all conditions for continuity. Hence
f (x) is a continuous function in R.
(iii) e x tan x
Let f(x) = e x tan x
f (x) is defined at ail points of R.
except at (2n + 1)\(\frac{\pi}{2}\), n ∈ Z.
Let x 0 be an arbitrary point in R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z
∴ Limit at x = x 0 exist and is equal to the value of the function f(x) at x = x 0.
Since x 0 is arbitrary the limit of the function. f(x) exists at all points in R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z and is equal to the value of the function f (x) at that points.
∴ f (x) satisfies all conditions for continuity. Hence,
f(x) is continuous at all points of R – (2n + 1)\(\frac{\pi}{2}\), n ∈ Z
(iv) e 2x + x 2
Let f(x) = e 2x + x 2
Clearly, f(x) is defined for all points in R.
Let x 0 be an arbitrary point in R.
From equations (1) and (2) we have,
The limit of the function f(x) exist at x = x 0 and is equal to the value of the function f(x) at x – x 0.
Since x 0 is an arbitrary point in R, the above is true for all points in R. Hence f (x) satisfies all conditions for continuity. Hence f (x) is continuous at all points of R.
(v) x. log x
Let f(x) = x log x
The function f(x) is defined in the open interval (0, ∞) since log x is defined for x > 0. Let x 0 be an arbitrary point in (0, ∞). Then
= x 0 log x 0
f(x 0 ) = x 0 log x 0
From equation (1) and (2) we have
∴ The limit of the function f(x) exists at x = x 0 and is equal to the value of the function.
Since x 0 is an arbitrary point the above is true for all points in (0, ∞).
∴ f(x) is continuous at all points of (0, ∞).
(vi) \(\frac{\sin x}{x^{2}}\)
f(x) = \(\frac{\sin x}{x^{2}}\)
f(x) is not defined at x = 0
∴ f(x) is defined for all points of R – {0}
Let x 0 be an arbitrary point in R – {0}. Then
From equation (1) and (2) we have
∴ The limit of the function f(x) exist at x = x 0 and is equal to the value of the function f(x) at x = x 0.
Since x 0 is an arbitrary point in R – {0}, the above result is true for all points in R – {0}.
∴ f(x) is continuous at all points of R – {0}.
(vii) \(\frac{x^{2}-16}{x+4}\)
Let f(x) = \(\frac{x^{2}-16}{x+4}\)
f(x) is not defined at x = – 4
∴ f(x) is defined for all points of R – {- 4}.
Let x 0 be an arbitrary point in R – {- 4}. Then
From equation (1) and (2) we have
Thus the limit of the function f (x) exist at x = x 0 and is equal to the value of the function f (x) at x = x 0.
Since x 0 is an arbitrary point in R – {- 4} the above result is true for all points in R – { – 4}.
∴ f(x) is continuous at all points of R – {- 4}.
(viii) |x + 2| + |x – 1|
let f(x) = |x + 2| + |x – 1|
f( x) is defined for all points of R. Let x 0 be an arbitrary point in R. Then
From equation (1) and (2) we get
Thus the limit of the function f(x) exist at x = x 0 and is equal to the value of the function at x = x 0. Since x = x 0 is an arbitrary point in R, the above
result is true for all points in R. Hence f (x) is continuous at all points of R.
(ix) \(\frac{|x-2|}{|x+1|}\)
Let f(x) = \(\frac{|x-2|}{|x+1|}\)
f(x) is defined for all points of R except at x = – 1.
∴ f (x) is defined for all points of R – { – 1 }.
Let x 0 be an arbitrary point in R – {- 1}.Then
From equation (1) and (2) we have
Hence the limit of the function f(x) at x = x 0 exists and is equal to the value of the function at x = x 0.
Since x = x 0 is an arbitrary point in R – { – 1 }, the above result is true for all points in R – {- 1).
∴ f(x) is continuous at all points of R – {- 1}.
(x) cot x + tan x
Let f(x) = cot x + tan x
∴ The limit of the function f(x) exists at x = x 0 and is equal to the value of the function f (x) at x = x 0.
Since x 0 is an arbitrary point, the above result is true for all points of R – \(\left\{\frac{\mathrm{n} \pi}{2}\right\}\), n ∈ z.
∴ f(x) is continuous at all points of R – \(\left\{\frac{\mathrm{n} \pi}{2}\right\}\), n ∈ Z

Clearly, the given function f(x) is defined at all points of R.
Case (i) Let x 0 ∈ (- ∞, 1) then
f (x) is continuous at x = x 0.
Since x 0 is arbitrary f(x) is continuous at all points of (-∞, 1).
Case (ii) Let x 0 ∈ (1, ∞) then
f (x) is continuous at x = x 0.
Since x 0 is an arbitrary point of (1, ∞), f(x) is continuous at all points of (1, ∞).
Case (iii) Let x 0 = 1 then
Hence, f (x) is continuous at x = 1.
Using all the three cases, we have f (x) is continuous at all the points of R.

When x < 0 We have y = 0
When 0 ≤ x < 2 We have y = x 2
When x ≥ 2 We have y = 4
Case (i) If x < 0 ie. (-∞, 0) then f (x) = 0
which is clearly continuous in (-∞, 0).
Case (ii) If 0 ≤ x < 2 je. [0, 2)
Let x 0 be an arbitrary point in [0, 2)
Hence f(x) is continuous at x = x 0. Since x = x 0 is an arbitrary f(x) is continuous at all points of [0, 2).
Case (iii) x ≥ 2 je. [2, ∞)
f( x) = 4 which is clearly continuous in [2, ∞)
Case (iv) at x = 2,
f(2) = 4
∴ f(x) is continuous at x = 2.
∴ Using case (j) case (ii) case (iii) and case (iv) we have f(x) is continuous at all points of R.

Given f and g are continuous functions.
2 f(3) – g(3) = 4
2 × 5 – g(3) = 4
10 – 4 = g(3)
g(3) = 6

Given g is continuous on R.
∴ g (x) is continuous at x = 4.
4 2 – b 2 = b × 4 + 20
16 – b 2 = 4b + 20
b 2 + 4b + 20 – 16 = 0
b 2 + 4b + 4 = 0
(b + 2) 2 = 0
b + 2 = 0 ⇒ b = -2

The function f (x) has a removable discontinuity at x = 1. Redefine f (x) as
∴ f(1) = \(\frac{2}{3}\). Then f(x) will be continuous at x = 1

(a)
The left – hand limit and right hand limit does not coincide at x = x 0
(b)
The function f(x) is not defined at x = x 0 and hence the continuity is destroyed at x = x 0
(c)
The limit of f(x) does not exist at x = x 0
(d)
The left hand limit and right – hand limit does not coincide at x = x 0

(2) 12
Explaination:
Given f(3) = 12
f takes only rational values
f(x) = 12
f(3) = 12
f(4.5) = 12