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Is the function f defined by f(x)=\left\{\begin{array}{lll}x_{,} & \text {if } & x \leq 1 \\ 5, & \text { if } & x>1\end{array}\right. continuous at x=0, at x=1, at x=2 ?
CBSE | Class 12 | Continuity and Differentiability | Exercise 5.1 | Maths | NCERT
Is the function f defined by f(x)=\left\{\begin{array}{lll}x_{,} & \text {if } & x \leq 1 \\ 5, & \text { if } & x>1\end{array}\right. continuous at x=0, at x=1, at x=2 ?

Solution:

The function provided is f(x)=\left\{\begin{array}{lll}x_{,} & \text {if } & x \leq 1 \\ 5, & \text { if } & x>1\end{array}\right.

Step 1: We all know that, f is defined at 0 and its value 0, at x=0.

Therefore \lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} x=0 and f(0)=0

As a result, f(x) is continuous at x=0.

Step 2: Left Hand limit (LHL) of f_{x \rightarrow 1}^{\lim _{x} f(x)=\lim _{x \rightarrow 1}(x)=1}, at x=1.

Right Hand limit (RHL) of f_{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1}(x)=5

Given here \lim _{x \rightarrow 1^{-}} f(x) \neq \lim _{x \rightarrow 1^{-}} f(x)

As a result, f(x) is not continuous at x=1.

Step 3: f is defined at 2 and its value at 2 is 5, at x=2.

\lim _{x \rightarrow 2} f(x)=\lim _{x \rightarrow 2}(5)=5, then, \lim _{x \rightarrow 2} f(x)=f(2)

As a result, f(x) is continuous at x=2.

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