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Question: How do you find the derivative of \( f(x) = 3 \) using the limit process?...

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

Explanation

Solution

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

Complete step-by-step answer :
Consider, the Given the function f(x)=3f(x) = 3
Step 1: Now, find the left-hand derivative of the function ff and states its derivative equals
LHD=lim⁡x→c−f(x)−f(c)x−cLHD = \mathop {\lim }\limits_{x \to {c^ - }} \dfrac{{f(x) - f(c)}}{{x - c}}
Here to examine the differentiability, take some substitution for lim⁡x→c−f(x)\mathop {\lim }\limits_{x \to {c^ - }} f(x) put x=c−hx = c - h and change the limit as x→c−x \to {c^ - } by h→0h \to 0 , then above equation becomes
⇒LHD=lim⁡h→0f(c−h)−f(c)−h\Rightarrow LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c - h) - f(c)}}{{ - h}}
So, when f(x)=3f(x) = 3 , we see that f(c−h)=3f(c - h) = 3 and f(c)=3f(c) = 3 as well, since 3 is a constant with no variable
⇒LHD=lim⁡h→03−3−h\Rightarrow LHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{3 - 3}}{{ - h}}
⇒LHD=lim⁡h→0  0−h\Rightarrow LHD = \mathop {\lim }\limits_{h \to 0} \,\,\dfrac{0}{{ - h}}
⇒LHD=lim⁡h→0 (0)=0\Rightarrow LHD = \mathop {\lim }\limits_{h \to 0} \,(0) = 0
Step 2: Next, find the right-hand derivative to the function ff and states its derivative equals RHD=lim⁡x→c+f(x)−f(c)x−cRHD = \mathop {\lim }\limits_{x \to {c^ + }} \dfrac{{f(x) - f(c)}}{{x - c}}
Here to examine the differentiability, take some substitution for lim⁡x→c+f(x)\mathop {\lim }\limits_{x \to {c^ + }} f(x) put x=c+hx = c + h and change the limit as x→c+x \to {c^ + } by h→0h \to 0 , then Equation (3) becomes
⇒RHD=lim⁡h→0f(c+h)−f(c)h\Rightarrow RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c + h) - f(c)}}{h}
So, when f(x)=3f(x) = 3 , we see that f(c+h)=3f(c + h) = 3 and f(c)=3f(c) = 3 as well, since 3 is a constant with no variable
⇒RHD=lim⁡h→03−3h\Rightarrow RHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{3 - 3}}{h}
⇒RHD =lim⁡h→00h\Rightarrow RHD\, = \mathop {\lim }\limits_{h \to 0} \dfrac{0}{h}
⇒RHD=lim⁡h→0 (0)=0\Rightarrow RHD = \mathop {\lim }\limits_{h \to 0} \,(0) = 0
By step 2 and 3 we get LHD=RHDLHD = RHD Hence the function ff is differentiable then the derivative of function ff exists
Step 3: The limit definition of the derivative takes a function ff and states its derivative equals
f′(x)=lim⁡h→0f(x+h)−f(x)hf'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}
So, when f(x)=3f(x) = 3 , we see that f(x+h)=3f(x + h) = 3 as well, since 3 is a constant with no variable
⇒f′(x)=lim⁡h→03−3h\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{3 - 3}}{h}
⇒f′(x)=lim⁡h→00h\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \dfrac{0}{h}
⇒f′(x)=lim⁡h→0(0)\Rightarrow f'(x) = \mathop {\lim }\limits_{h \to 0} \left( 0 \right)
⇒f′(x)=0\Rightarrow f'(x) = 0
Hence the derivative of f(x)=3f(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=lim⁡h→0f(c−h)−f(c)−hLHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c - h) - f(c)}}{{ - h}} and the formula for right hand derivative is RHD=lim⁡h→0f(c+h)−f(c)hRHD = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(c + h) - f(c)}}{h}