Q.Examine the differentiability of $f$, where $f$ is defined by $f(x) = \begin{cases} x^2 \sin \dfrac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ at $x = 0$.

Concept: Differentiability at a Point (First Principles) — since $f$'s formula involves $\sin(1/x)$, we test differentiability directly from the limit definition, not by ordinary differentiation rules.

Step 1: $\dfrac{f(0+h)-f(0)}{h}=\dfrac{h^2\sin(1/h)-0}{h}=h\sin(1/h)$, for $h \neq 0$. …

Why First Principles Is Needed Here

$f$'s formula changes at $x=0$ ($x^2\sin(1/x)$ for $x\ne0$, then $0$ at $x=0$), and the factor $\sin(1/x)$ oscillates without settling as $x\to0$. Ordinary differentiation rules can't be trusted right at the join, so we go back to the limit definition of the derivative.

Step 1 — Write the difference quotient

Step 2 — Bound the oscillating factor

Since $-1\le\sin(1/h)\le1$ for every $h\ne0$, multiplying by $|h|$ gives

Step 3 — Apply the Squeeze Theorem …

Method: Differentiability at a Point via the Squeeze Theorem for an Oscillating Function

This method applies when the difference quotient at a point involves a bounded but wildly oscillating factor (like $\sin\frac{1}{h}$ or $\cos\frac{1}{h}$) multiplied by a factor that shrinks to zero.

Steps

Step 1: Form the difference quotient directly from the definition

Substitute the piece of the function valid near $a$ (excluding $a$ itself) and simplify algebraically — cancelling common factors of $h$ where possible.

Step 2: Identify the bounded oscillating part and isolate it

If the simplified quotient contains a term like $\sin\left(\frac{1}{h}\right)$ or $\cos\left(\frac{1}{h}\right)$, recognise that such a term is always trapped between $-1$ and $1$ for every $h \ne 0$, however wildly it oscillates as $h \to 0$.

Step 3: Apply the squeeze (sandwich) theorem

Bound the whole expression between two simpler functions that both tend to the same limit:

$-|h| \le h\sin\left(\frac{1}{h}\right) \le |h|$ …

Common Mistakes

Mistake 1: Assuming the limit doesn't exist because $\sin\frac1h$ oscillates

Why it's wrong: students often see the unbounded oscillation of $\sin(1/h)$ as $h\to0$ and conclude the whole limit is undefined — but they've ignored that this oscillating factor is being multiplied by a term (here, $h$) that shrinks to zero, which forces the product to zero regardless of the oscillation. Correct approach: whenever a bounded oscillating factor is multiplied by something tending to zero, apply the squeeze theorem rather than concluding the limit fails to exist.

Mistake 2: Concluding the derivative is continuous just because it exists at the point …