Q.Prove that the function $f$ defined by $f(x) = \begin{cases} \dfrac{x}{|x| + 2x^2}, & x \ne 0 \\ k, & x = 0 \end{cases}$ remains discontinuous at $x = 0$, regardless the choice of $k$.

Concept: Continuity at a point — if the one-sided limits differ, no value of $k$ can restore continuity.

Left of $0$ ($x<0$, so $|x|=-x$):

Right of $0$ ($x>0$, so $|x|=x$):

$f(x)=\frac{x}{x+2x^2}=\frac{x}{x(1+2x)}=\frac{1}{1+2x}\xrightarrow[x\to 0^+]{}1.$ …

Why the choice of $k$ can't help

$k$ only sets the single value $f(0)$. Continuity needs the limit to exist and equal that value. If the left and right limits disagree, the limit itself fails to exist, and then there is nothing $f(0)$ can match — $k$ is powerless.

The absolute value $|x|$ makes $f$ behave differently on the two sides of $0$, so we examine each side.

Step-by-step

1. Right side ($x>0$). Here $|x|=x$, so

Cancelling $x$ is legal because $x\ne 0$. As $x\to 0^+$, $1+2x\to 1$, so

2. Left side ($x<0$). Here $|x|=-x$, so

As $x\to 0^-$, $-1+2x\to -1$, so

$\lim_{x\to 0^-}f(x)=-1.$ …

Method: Proving Discontinuity Is Unavoidable for Every Value of a Parameter

This method applies when a problem asks you to show that no choice of a constant $k$ (used to define $f(a)$) can make a piecewise function continuous at $a$ — the constant only controls the function's value at the point, not the limit approaching it.

Steps

Step 1: Recognise what the constant can and cannot fix

A constant like $k$ defined as $f(a)=k$ only sets the single value at the point. It cannot change what the function does on either side of $a$. So continuity is only possible if $\displaystyle\lim_{x\to a}f(x)$ exists in the first place — if it doesn't, no $k$ can rescue continuity.

Step 2: Split the expression by sign and simplify each branch …

Common Mistakes

Mistake 1: Sign error when replacing $|x|$ on the two sides of the point

Why it's wrong: students often use $|x|=x$ on both sides, or forget that $|x|=-x$ for $x<0$ — this silently changes the algebra and can make a genuine jump discontinuity look removable. Correct approach: explicitly write $|x|=-x$ for $x<0$ and $|x|=x$ for $x>0$ before simplifying, treating the two cases completely separately.

Mistake 2: Believing the constant $k$ could still "fix" a jump discontinuity …