NTA Abhyas 2020
- Exam
- NTA Abhyas
- Year
- 2020
- Subject
- Mathematics
- Chapter
- Continuity and Differentiability
NTA Abhyas 2020 · Mathematics · Continuity and Differentiability
The function f(x) = _ n → ∞ (x - 2) ^ 2n - 1 (x - 2) ^ 2n + 1 ( n ∈...
The function f\left(x\right) = \lim_{n \to \infty}\dfrac{{ \left(x - 2\right)}^{2n} - 1}{{ \left(x - 2\right)}^{2n} + 1} \left(\forall n \in \mathbb{N}\right) is discontinuous at&
Options
- A only
- B only
- C and
- D and
Correct answer
C. and
Step-by-step solution
Given,
\left\{\right)
⇒ f\left(x\right) = \left\{\begin{cases}1, & x 3\end{cases}\right)<br />Thus,f \left(x\right)x = 1,3
\left\{\right)
⇒ f\left(x\right) = \left\{\begin{cases}1, & x 3\end{cases}\right)<br />Thus,f \left(x\right)x = 1,3