Question

How do you find the derivative of $ f(x) = 3 $ using the limit process?

Answer 1

Hint: A real function $ f $ is said to be derivable or differentiable at a point $ c $ in its domain, if its left hand derivative and right hand derivatives at $ c $ exists it will be a finite and unique and both are should be equal (i.e., $ LHD = RHD $ or $ Lf'(c) = Rf'(c) $ ), otherwise the function $ f $ is not differentiable then derivative not exist.

Complete step-by-step answer:
Consider, the Given the function $ f(x) = 3 $
Step 1: Now, find the left-hand derivative of the function $ f $ and states its derivative equals
 $ LHD = \mathop {\lim }\limits_{x \to {c^ - }} \dfrac{{f(x) - f(c)}}{{x - c}} $
Here to examine the differentiability, take some substitution for $ \mathop {\lim }\limits_{x \to {c^ - }} f(x) $ put $ x = c - h $ and change the limit as $ x \to {c^ - } $ by $ h \to 0 $ , then above equation becomes
 $ \Rightarrow LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c - h) - f(c)}}{{ - h}} $
So, when $ f(x) = 3 $ , we see that $ f(c - h) = 3 $ and $ f(c) = 3 $ as well, since 3 is a constant with no variable
 $ \Rightarrow LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{3 - 3}}{{ - h}} $
  $ \Rightarrow LHD = \mathop {\lim }\limits_{h \to 0} \,\,\dfrac{0}{{ - h}} $
 $ \Rightarrow LHD = \mathop {\lim }\limits_{h \to 0} \,(0) = 0 $
Step 2: Next, find the right-hand derivative to the function $ f $ and states its derivative equals $ RHD = \mathop {\lim }\limits_{x \to {c^ + }} \dfrac{{f(x) - f(c)}}{{x - c}} $
Here to examine the differentiability, take some substitution for $ \mathop {\lim }\limits_{x \to {c^ + }} f(x) $ put $ x = c + h $ and change the limit as $ x \to {c^ + } $ by $ h \to 0 $ , then Equation (3) becomes
 $ \Rightarrow RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c + h) - f(c)}}{h} $
So, when $ f(x) = 3 $ , we see that $ f(c + h) = 3 $ and $ f(c) = 3 $ as well, since 3 is a constant with no variable
 $ \Rightarrow RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{3 - 3}}{h} $
  $ \Rightarrow RHD\, = \mathop {\lim }\limits_{h \to 0} \dfrac{0}{h} $
 $ \Rightarrow RHD = \mathop {\lim }\limits_{h \to 0} \,(0) = 0 $
By step 2 and 3 we get $ LHD = RHD $ Hence the function $ f $ is differentiable then the derivative of function $ f $ exists
Step 3: The limit definition of the derivative takes a function $ f $ and states its derivative equals
 $ f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h} $
So, when $ f(x) = 3 $ , we see that $ f(x + h) = 3 $ as well, since 3 is a constant with no variable
 $ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{3 - 3}}{h} $
 $ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{0}{h} $
 $ \Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( 0 \right) $
 $ \Rightarrow f'(x) = 0 $
 Hence the derivative of $ f(x) = 3 $ is 0.
So, the correct answer is “0”.

Note: The function is differentiable and then the function should exist limit. In the limit process we have to find the first left-hand derivative and right-hand derivative. If the left-hand derivative is equal to the right hand derivative, then it is differentiable. The formula for left hand derivative is $ LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c - h) - f(c)}}{{ - h}} $ and the formula for right hand derivative is $ RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c + h) - f(c)}}{h} $