Animated Solution for Mathematics - Limits, Continuity and Differentiability: If $f(x)$ is continuous and differentiable function and $f(1/n)=0$ for all $n\ge 1,n\in I$, then

Imagine you are standing on the $x$-axis, looking at a function $f(x)$ that is continuous and differentiable. You are told something peculiar: at every point $x=\frac{1}{n}$ (where $n$ is a positive integer), the function value is exactly zero.

That is, $f(1)=0$, $f(1/2)=0$, $f(1/3)=0$, and so on. It feels like the function is playing a game of 'hide and seek' with the $x$-axis, popping up to touch it at infinitely many points.

But what does this tell us about the function at the origin, $x=0$? Let us embark on a journey to uncover the truth.

First, let us visualize the points $x_n=\frac{1}{n}$. As $n$ grows larger and larger, the value of $\frac{1}{n}$ gets smaller and smaller.

The sequence of points $1,1/2,1/3,1/4,\dots$ is marching steadily toward the origin. They are crowding infinitely close to $x=0$.

This is the crucial observation. We are not just looking at a few points; we are looking at an infinite sequence of roots that accumulate at the origin.

We are given that $f(x)$ is continuous everywhere. This is a powerful piece of information.

Continuity at $x=0$ means that the function value at zero must be equal to the limit of the function as we approach zero from any direction. Mathematically, $f(0)={\lim }_{x\to 0}f(x)$.

Since we have a sequence of points $x_n=\frac{1}{n}$ that approaches zero, we can evaluate the limit along this sequence:

Since we know $f(\frac{1}{n})=0$ for all $n$, the limit is simply the limit of zero, which is zero. Thus, we have firmly established that $f(0)=0$. The function is anchored to the origin.

Now, let us tackle the slope. The problem states that $f(x)$ is differentiable. This means the derivative $f^{\prime }(0)$ exists.

By the first principle of derivatives, the slope at $x=0$ is defined as:

We need to evaluate this limit as $h$ approaches zero. We have the perfect tool: our sequence $h=\frac{1}{n}$. As $n\to \infty$, $h\to 0$.

Let us substitute this into our derivative definition:

We already know that $f(\frac{1}{n})=0$ and $f(0)=0$. Substituting these values, the expression becomes:

For any finite $n$, the denominator $\frac{1}{n}$ is a non-zero number. Zero divided by any non-zero number is zero.

Therefore, we are taking the limit of a constant zero sequence. The result is $f^{\prime }(0)=0$. The function is not just passing through the origin; it is doing so with a perfectly horizontal tangent.

It is tempting to think that if a function is zero at so many points, it must be the zero function everywhere. But that is the beauty of calculus!

A function can be zero at infinitely many points and still be non-zero elsewhere. Consider the function $f(x)=x^2\sin (\frac{\pi }{x})$ for $x eq 0$ and $f(0)=0$.

This function oscillates between positive and negative values, hitting zero at $x=\frac{1}{n}$, yet it remains continuous and differentiable at the origin. This problem teaches us that local behavior at a point can be constrained by a sequence of points, even if the global behavior is complex.