Answer
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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.
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