Q.Find the value of $k$ so that the function $f$ is continuous at the indicated point: $f(x) = \begin{cases} \dfrac{2^{x+2} - 16}{4^x - 16}, & x \ne 2 \\ k, & x = 2 \end{cases}$ at $x = 2$.

Concept: Continuity At A Point — for continuity, the value $k = f(2)$ must equal $\displaystyle\lim_{x \to 2} f(x)$.

Step 1: At $x=2$: numerator $2^{2+2}-16=0$, denominator $4^2-16=0$ — an indeterminate $\frac00$ form.

Step 2: Let $u = 2^x$. Then $\dfrac{4u-16}{u^2-16} = \dfrac{4(u-4)}{(u-4)(u+4)} = \dfrac{4}{u+4}$ for $u \neq 4$. …

Setting Up

$f$ is continuous at $x=2$ exactly when $\displaystyle\lim_{x\to2}f(x)=f(2)=k$. Plugging $x=2$ directly into the rational expression gives $\frac{0}{0}$ (an indeterminate form), so we factor to find the hidden cancelling term.

Step 1 — Rewrite in terms of $2^x$

Note $2^{x+2}=4\cdot2^x$ and $4^x=(2^2)^x=2^{2x}=(2^x)^2$. Let $u=2^x$:

Step 2 — Cancel the common factor

For $x\ne2$, $u=2^x\ne4$, so we may cancel $(u-4)$:

Step 3 — Take the limit

$\lim_{x\to2}f(x)=\lim_{x\to2}\frac{4}{2^x+4}=\frac{4}{2^2+4}=\frac{4}{8}=\frac12.$ …

Method: Finding an Unknown Constant via an Indeterminate-Form Exponential Limit

This method applies when the unknown constant $k$ must be chosen to fill a removable discontinuity in an exponential expression — i.e. the limit as $x \to a$ of the given ("$x \ne a$") branch exists, and $k$ must equal that limit for continuity.

Steps

Step 1: Substitute $x=a$ directly to confirm the indeterminate form.

If both the numerator and denominator vanish (or both blow up), a genuine limit — not a simple substitution — is needed, and $k$ must be set equal to whatever that limit turns out to be.

Step 2: Rewrite every exponential term as a power of a single common base.

Use index laws such as $b^{m+n} = b^m \cdot b^n$ and $(b^m)^n = b^{mn}$ to express every term (numerator and denominator alike) as powers of the same base raised to $x$ — this reveals the hidden algebraic structure.

Step 3: Treat the common-base power as a single variable (e.g. let $t = b^x$) and factor. …

Common Mistakes

Mistake 1: Reaching for L'Hopital's Rule instead of factoring.

Why it's wrong: L'Hopital's Rule is outside the CBSE Class 12 syllabus for this chapter, and reaching for it here also obscures the underlying algebraic structure (a difference-of-squares factorization) that the problem is testing. Correct approach: rewrite every exponential term as a power of the same base (here, base $2$) and factor algebraically — the cancellation reveals the limit directly.

Mistake 2: Misapplying the index laws while rewriting $4^x$ and $2^{x+2}$ in terms of $2^x$. …