How do you find the points of continuity of a function?
Answer
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Hint: We first discuss the concept of the continuity for a function $f\left( x \right)$ at a particular point $x=a$. We try to find the limiting value. We also use examples to understand the concept better. We find the relation between continuity and the differentiation.
Complete step by step answer:
A given function $f\left( x \right)$ is continuous if the limiting value of the function at a particular point is equal from both ends.
This means if we have to check the continuity of the function $f\left( x \right)$ at point $x=a$ then we have to find the value of the function at three parts $x={{a}^{+}},{{a}^{-}},a$.
If the equation $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ holds then we can say that the function is continuous at $x=a$.
We take two functions to understand the theorem better.
Let $f\left( x \right)=\left| x \right|$ and we find continuity at $x=0$.
Now $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,x=0$, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( -x \right)=0$ and $f\left( 0 \right)=0$.
Therefore, $f\left( x \right)=\left| x \right|$ is continuous at $x=0$.
Now we take $f\left( x \right)=\dfrac{1}{x}$ and we find continuity at $x=0$.
Now $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1}{x}=+\infty $, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{1}{x}=-\infty $ and $f\left( 0 \right)=\text{undefined}$.
Therefore, $f\left( x \right)=\dfrac{1}{x}$ is not continuous at $x=0$.
Note: The differentiation of a function is connected to its continuity where if a function is differentiable then it is definitely continuous. But the opposite is not always true. A function being continuous doesn’t make it differentiable.
Complete step by step answer:
A given function $f\left( x \right)$ is continuous if the limiting value of the function at a particular point is equal from both ends.
This means if we have to check the continuity of the function $f\left( x \right)$ at point $x=a$ then we have to find the value of the function at three parts $x={{a}^{+}},{{a}^{-}},a$.
If the equation $\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,f\left( x \right)=f\left( a \right)$ holds then we can say that the function is continuous at $x=a$.
We take two functions to understand the theorem better.
Let $f\left( x \right)=\left| x \right|$ and we find continuity at $x=0$.
Now $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,x=0$, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,f\left( x \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( -x \right)=0$ and $f\left( 0 \right)=0$.
Therefore, $f\left( x \right)=\left| x \right|$ is continuous at $x=0$.
Now we take $f\left( x \right)=\dfrac{1}{x}$ and we find continuity at $x=0$.
Now $\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\dfrac{1}{x}=+\infty $, $\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\dfrac{1}{x}=-\infty $ and $f\left( 0 \right)=\text{undefined}$.
Therefore, $f\left( x \right)=\dfrac{1}{x}$ is not continuous at $x=0$.
Note: The differentiation of a function is connected to its continuity where if a function is differentiable then it is definitely continuous. But the opposite is not always true. A function being continuous doesn’t make it differentiable.
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