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# Prove that $\left| x \right|$ is not differentiable at $x=0$ by using first principle?

Last updated date: 20th Jun 2024
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Hint: These types of problems are pretty straight forward and are very easy to solve. For problems like these we need to remember all the concepts of the theory of limits including the first principle. According to the first principle of limits, say we have a function $f\left( x \right)$ and we consider a point on this curve as $\left( x,f\left( x \right) \right)$ and another point $\left( x+h,f\left( x+h \right) \right)$ where $h$ is an infinitesimal quantity, then the derivative of the function $f\left( x \right)$ is defined as,
${{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}$
Now, for any limit value to exist, both the value of the left hand limit and the right hand limit must be equal.

Now, we start off the solution to the given problem by writing that,
We consider the given function$f\left( x \right)=\left| x \right|$. Using the first principle of derivatives, we can write,
${{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{\left| x+h \right|-\left| x \right|}{h}$
Now, considering the right hand limit, we consider, $x\to {{0}^{+}}$, which means $x$ is approaching $0$ from the right hand side in the number line. Hence we can write,
\begin{align} & {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{x+h-x}{h} \\ & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{h}{h} \\ \end{align}
Evaluating the value of the right hand limit we get,
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=1$
Now, considering the left hand limit, we consider, $x\to {{0}^{-}}$, which means $x$ is approaching $0$ from the left hand side in the number line. Hence we can write,
\begin{align} & {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{-\left( x+h \right)-\left( -\left( x \right) \right)}{h} \\ & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{-x-h+x}{h} \\ & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=\displaystyle \lim_{h \to 0}\dfrac{-h}{h} \\ \end{align}
Evaluating the value of the right hand limit we get,
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{dy}{dx}=-1$
Thus from the above evaluations, we get the value of the right hand limit as $1$ and the left hand limit as $-1$ . Since the value of the right hand and left limits are not equal, we say that this limit does not exist.

Note: For problems like these, we must remember the theory of limits as well as the definition of the first principle for finding derivatives of a function. We can also solve this particular problem in another method, i.e. by using the theory of graphs. If we plot the graph of $f\left( x \right)=\left| x \right|$ we will see that at point $x=0$ there is a sharp point edge, and for such pointed edges, the function is non-differentiable. We can apply this concept for any given function to find out the points of non-differentiability very quickly and easily.