Q.Find whether the function is continuous or discontinuous at the indicated point: $f(x) = \begin{cases} |x| \cos \dfrac{1}{x}, & x \ne 0 \\ 0, & x = 0 \end{cases}$ at $x = 0$.

Concept: Continuity At A Point — A function $f$ is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

Step 1: We need $\lim_{x \to 0} f(x)$. For $x \ne 0$, $f(x) = |x| \cos \frac{1}{x}$.

Step 2: Since $|\cos \frac{1}{x}| \le 1$, we have $0 \le |f(x)| = |x| \cdot |\cos \frac{1}{x}| \le |x|$.

Step 3: As $x \to 0$, $|x| \to 0$. By the Sandwich Theorem, $\lim_{x \to 0} |f(x)| = 0$, so $\lim_{x \to 0} f(x) = 0$. …

The Concept: Continuity at a Point

For a function to be continuous at $x = a$, three things must hold:

1. $f(a)$ is defined.
2. $\lim_{x \to a} f(x)$ exists.
3. $\lim_{x \to a} f(x) = f(a)$.

Here, $f(0) = 0$ is given, so condition 1 is satisfied. The real question is whether the limit exists and equals zero. The function $|x| \cos(1/x)$ is a classic example of the Squeeze Theorem in action — the cosine part oscillates wildly near zero, but it's trapped between $-1$ and $1$, while $|x|$ calmly marches to zero.

Step-by-Step Solution

1. Set up the limit we need to check

   We want $\displaystyle \lim_{x \to 0} f(x) = \lim_{x \to 0} |x| \cos\left(\frac{1}{x}\right)$.

   The function is defined piecewise, but for $x \neq 0$, it's just $|x| \cos(1/x)$.

2. Bound the oscillating part

   For any real $t$, we know $-1 \leq \cos t \leq 1$. So for any $x \neq 0$:

3. Multiply by $|x|$ (which is always non-negative) Multiplying the inequality by $|x|$ preserves the direction: $-|x| \leq |x| \cos\left(\frac{1}{x}\right) \leq |x|$ …

Method: Using the Squeeze (Sandwich) Theorem for a Bounded-Times-Vanishing Product

This method applies whenever the expression whose limit is needed is a product of two factors: one factor that provably tends to $0$, and another factor that is bounded (even though it oscillates and has no limit of its own) — a very common construction at the boundary of a piecewise function.

Steps

Step 1: Identify the two factors.

Separate the expression into a "shrinking" factor (e.g. $|x|$ or $x^2$, which tends to $0$) and an "oscillating but bounded" factor (e.g. $\cos\frac1x$ or $\sin\frac1x$), which itself has no limit as $x \to 0$.

Step 2: State the boundedness of the oscillating factor explicitly.

Every trig function satisfies $-1 \le \cos(\cdot) \le 1$ and $-1 \le \sin(\cdot) \le 1$, no matter how wild its argument is. Write this as an inequality involving the actual oscillating factor in the problem. …

Common Mistakes

Mistake 1: Concluding the whole limit does not exist because $\cos\frac1x$ itself has no limit.

Why it's wrong: $\cos\frac1x$ indeed oscillates without settling on a value, but it's multiplied by $|x|$, which tends to $0$; a bounded oscillation times a vanishing factor still has a well-defined limit (zero), even though the oscillating factor alone does not. Correct approach: recognise the "bounded $\times$ vanishing" structure and apply the Squeeze Theorem rather than reasoning about the oscillating factor in isolation.

Mistake 2: Multiplying the bounding inequality by a possibly-negative quantity without adjusting the inequality direction. …