The function fx=limn→∞(x-2^2n-1)/(x-2^2n+1) ∀n∈N is discontinuous at

Correct answer: (C) x=1 and 3. Given, fx=limn→∞(x-2^2)/(n-1x-2^2n+1) ⇒fx=limn→∞(1-1x-2^2n)/(1+1x-2^2n) ⇒fx= -1, 0, 1, 0≤x-2^2 1 ⇒fx= 1, x 3 Thus, fx is discontinuous at x=1,3

Q10 / 93 · NTA Abhyas (JEE Main) 2020 · Mathematics · Continuity and Differentiability

The function $f (x) = \underset{n \to \infty}{l i m} \frac{{(x - 2)}^{2 n} - 1}{{(x - 2)}^{2 n} + 1}$ $(\forall n \in N)$ is discontinuous at

A
$x = 1$ only
B
$x = 3$ only
Correct answer
C
$x = 1$ and $3$
D
$x = 0,1$ and $2$

Solution

Given,

f (x) = \underset{n \to \infty}{l i m} \frac{{[{(x - 2)}^{2}]}^{n} - 1}{{[{(x - 2)}^{2}]}^{n} + 1}

\Rightarrow f (x) = \underset{n \to \infty}{l i m} \frac{1 - \frac{1}{{[{(x - 2)}^{2}]}^{n}}}{1 + \frac{1}{{[{(x - 2)}^{2}]}^{n}}}

\Rightarrow f (x) = \{\begin{array}{l} - 1, \\ 0, \\ 1, \end{array} \begin{array}{l} 0 \leq {(x - 2)}^{2} < 1 \\ {(x - 2)}^{2} = 1 \\ {(x - 2)}^{2} > 1 \end{array}

\Rightarrow f (x) = \{\begin{cases} 1, & x < 1 \\ 0, & x = 1 \\ - 1, & 1 < x < 3 \\ 0, & x = 3 \\ 1, & x > 3 \end{cases}

Thus,

f (x)

is discontinuous at

x = 1,3

Hence, the correct answer is option: 'C'.

The function fx=limn→∞x-22n-1x-22n+1 ∀n∈N is discontinuous at

Solution

More Continuity and Differentiability Questions

The function $f (x) = \underset{n \to \infty}{l i m} \frac{{(x - 2)}^{2 n} - 1}{{(x - 2)}^{2 n} + 1}$ $(\forall n \in N)$ is discontinuous at