Show that every differentiable function is continuous (converse is not true i.e., a function may be continuous but not differentiable).
Answer
548k+ views
Hint: Here we will use the basic definition of the differential function. Then we will form the condition of a continuous function. We will then show that the function is continuous to prove that every differentiable function is a continuous function.
Complete step-by-step answer:
Let \[f\] be the differentiable function at \[x = a\].
Then according to the basic definition of the differentiation, Differentiation of a function is equals to
\[f'\left( c \right) = \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\]……………………….\[\left( 1 \right)\]
We know that the for a function to be continuous at a point it must satisfy the equation
\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\]
We can write the above equation as
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0\]……………………\[\left( 2 \right)\]
So, for a function to be continuous it must satisfy the equation \[\left( 2 \right)\].
Now we will find the value of \[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right)\] for the given differentiable function.
Therefore we can write \[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right)\] as,
\[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\left( {x - a} \right)} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \right) \times \mathop {\lim }\limits_{x \to a} \left( {x - a} \right)\]
By using the equation \[\left( 1 \right)\] in the above equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times \mathop {\lim }\limits_{x \to a} \left( {x - a} \right)\]
By putting the limit on the RHS of the equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times \left( {a - a} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times 0\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0\]
Hence as per the condition of the equation \[\left( 2 \right)\] we can say that the given function \[f\] is a continuous function.
Hence, every differentiable function is continuous.
Note: Here we have to note that continuous function is the function whose value does not change or value remains constant. When the function is continuous at a point then the left hand limit of the function and the right hand limit of the function is equal to the value of the function at that point.
\[\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)\]
Also, a differentiable function is always continuous but the converse is not true which means a function may be continuous but not always differentiable. A differentiable function may be defined as is a function whose derivative exists at every point in its range of domain.
Complete step-by-step answer:
Let \[f\] be the differentiable function at \[x = a\].
Then according to the basic definition of the differentiation, Differentiation of a function is equals to
\[f'\left( c \right) = \mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\]……………………….\[\left( 1 \right)\]
We know that the for a function to be continuous at a point it must satisfy the equation
\[\mathop {\lim }\limits_{x \to a} f\left( x \right) = f\left( a \right)\]
We can write the above equation as
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0\]……………………\[\left( 2 \right)\]
So, for a function to be continuous it must satisfy the equation \[\left( 2 \right)\].
Now we will find the value of \[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right)\] for the given differentiable function.
Therefore we can write \[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right)\] as,
\[\mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}\left( {x - a} \right)} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = \mathop {\lim }\limits_{x \to a} \left( {\dfrac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \right) \times \mathop {\lim }\limits_{x \to a} \left( {x - a} \right)\]
By using the equation \[\left( 1 \right)\] in the above equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times \mathop {\lim }\limits_{x \to a} \left( {x - a} \right)\]
By putting the limit on the RHS of the equation, we get
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times \left( {a - a} \right)\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = f'\left( c \right) \times 0\]
\[ \Rightarrow \mathop {\lim }\limits_{x \to a} \left( {f\left( x \right) - f\left( a \right)} \right) = 0\]
Hence as per the condition of the equation \[\left( 2 \right)\] we can say that the given function \[f\] is a continuous function.
Hence, every differentiable function is continuous.
Note: Here we have to note that continuous function is the function whose value does not change or value remains constant. When the function is continuous at a point then the left hand limit of the function and the right hand limit of the function is equal to the value of the function at that point.
\[\mathop {\lim }\limits_{x \to {a^ - }} f(x) = \mathop {\lim }\limits_{x \to {a^ + }} f(x) = f(a)\]
Also, a differentiable function is always continuous but the converse is not true which means a function may be continuous but not always differentiable. A differentiable function may be defined as is a function whose derivative exists at every point in its range of domain.
Recently Updated Pages
Master Class 12 Business Studies: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Master Class 12 Biology: Engaging Questions & Answers for Success

Class 12 Question and Answer - Your Ultimate Solutions Guide

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
Which is more stable and why class 12 chemistry CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

What are the major means of transport Explain each class 12 social science CBSE

Sulphuric acid is known as the king of acids State class 12 chemistry CBSE

