Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $f(x)=\{\begin{array}{ll}{x}^2\cos \frac{\pi }{x}& x\ne 0\\ 0& x=0\end{array},x\in R$ then $f$ is

Welcome, fellow explorers of calculus. Today, we are diving into a problem that perfectly illustrates the difference between "smooth" behavior and the subtle, hidden truths of limits.

We are examining the function $f(x)=x^2\cos \left(\frac{\pi }{x}\right)$ for $x eq 0$, with $f(0)=0$. Our goal is to determine if this function is differentiable at two very different points: the origin ($x=0$) and a point in the wild ($x=2$).

The Origin and the First Principle

When we stand at $x=0$, we are at a boundary. The function is defined by a specific rule at this point, but by a completely different algebraic expression everywhere else.

Because of this, we cannot simply take the derivative of the expression $x^2\cos \left(\frac{\pi }{x}\right)$ and plug in $0$. Instead, we must return to the bedrock of calculus: the First Principle of Derivatives.

The definition of the derivative at $x=0$ is given by the limit:

Substituting our function, we get:

Simplifying this, we find ourselves staring at the limit:

The Squeeze Theorem to the Rescue

This is where the problem gets thrilling. As $h$ approaches $0$, the term $\frac{\pi }{h}$ explodes toward infinity. The cosine of an infinite angle oscillates wildly between $-1$ and $1$.

If you were to graph this, you would see the function frantically bouncing between the parabolic envelopes $y=x^2$ and $y=-x^2$. Does this mean the derivative doesn't exist? Not necessarily!

We know that for any value of $h$:

If we multiply this entire inequality by $|h|$, we get:

As $h$ approaches $0$, both $-|h|$ and $|h|$ shrink to $0$. By the Squeeze Theorem, the expression trapped in the middle is forced to $0$.

Thus, $f^{\prime }(0)=0$. The function is differentiable at the origin, and its tangent is a flat, horizontal line!

The Victory Lap at $x=2$

Now that we have conquered the origin, $x=2$ feels like a walk in the park. We are far from the piecewise boundary, so we can use our standard toolkit.

Our function is a product of two functions: $u=x^2$ and $v=\cos \left(\frac{\pi }{x}\right)$. We use the Product Rule: $f^{\prime }(x)=u^{\prime }v+u{v}^{\prime }$.

The derivative of $x^2$ is $2x$. For the derivative of $\cos \left(\frac{\pi }{x}\right)$, we use the Chain Rule:

Putting it all together, we get:

The $x^2$ terms cancel out, leaving:

Evaluating this at $x=2$:

Since $\cos \left(\frac{\pi }{2}\right)=0$ and $\sin \left(\frac{\pi }{2}\right)=1$, the first term vanishes, and we are left with $f^{\prime }(2)=\pi$. The function is perfectly differentiable here as well.

We have successfully navigated the treacherous waters of the origin and the smooth terrain of $x=2$. We found that the function is differentiable at both points.

This problem is a beautiful reminder that in calculus, even when things look chaotic, there is often a rigorous, elegant structure waiting to be revealed if you just apply the right tools.