Q.Let $f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2 \sin t & t & 2t \\ \sin t & t & t \end{vmatrix}$, then $\displaystyle\lim_{t \to 0} \dfrac{f(t)}{t^2}$ is equal to
(A) $0$
(B) $-1$
(C) $2$
(D) $3$

Concept: Determinant Range Analysis — expand the determinant, simplify using small‑angle approximations, then evaluate the limit.

First, expand $f(t)$ along the first row:

Compute each $2\times2$ determinant:

• $\begin{vmatrix} t & 2t \\ t & t \end{vmatrix} = t\cdot t - 2t\cdot t = t^2 - 2t^2 = -t^2$.
• $\begin{vmatrix} 2\sin t & 2t \\ \sin t & t \end{vmatrix} = (2\sin t)(t) - (2t)(\sin t) = 2t\sin t - 2t\sin t = 0$.
• $\begin{vmatrix} 2\sin t & t \\ \sin t & t \end{vmatrix} = (2\sin t)(t) - (t)(\sin t) = 2t\sin t - t\sin t = t\sin t$.

Thus $f(t) = \cos t \cdot (-t^2) + 0 + 1 \cdot (t\sin t) = -t^2\cos t + t\sin t$.

Now use small‑angle expansions: $\cos t = 1 - \frac{t^2}{2} + O(t^4)$, $\sin t = t - \frac{t^3}{6} + O(t^5)$. …

We need the limit of a determinant divided by $t^2$ as $t \to 0$. The determinant itself vanishes at $t = 0$ (check: all entries become constants, and the first and third rows become identical), so we have a $\frac{0}{0}$ form. That suggests using series expansions or L'Hôpital's rule. But the neatest path is to expand the determinant directly, then expand each term as a power series in $t$, and see what the leading behaviour is.

Why determinant range analysis works here:

The determinant is a polynomial (actually a combination of trigonometric and polynomial terms) in $t$. By expanding it, we get a sum of products of entries. Each product is of the form (some trig function) × (some polynomial in $t$). Expanding each trig function as a Maclaurin series lets us collect the lowest power of $t$ in the whole determinant. Dividing by $t^2$ and taking the limit then just picks off the coefficient of $t^2$ in the expansion of $f(t)$.

Let's do it step by step.

1. Expand the determinant along the first row (or any row — the result is the same). Using the first row:

2. Compute each 2×2 determinant:

• First minor: $\begin{vmatrix} t & 2t \\ t & t \end{vmatrix} = t \cdot t - 2t \cdot t = t^2 - 2t^2 = -t^2$.

• Second minor: $\begin{vmatrix} 2\sin t & 2t \\ \sin t & t \end{vmatrix} = (2\sin t)(t) - (2t)(\sin t) = 2t\sin t - 2t\sin t = 0$.

• Third minor: $\begin{vmatrix} 2\sin t & t \\ \sin t & t \end{vmatrix} = (2\sin t)(t) - (t)(\sin t) = 2t\sin t - t\sin t = t\sin t$.

So the expansion simplifies dramatically:

$f(t) = (\cos t)(-t^2) - t \cdot 0 + 1 \cdot (t\sin t) = -t^2 \cos t + t \sin t.$ …

Method: Evaluating a 0/0 Limit Involving a Determinant

When a limit involves a determinant that vanishes at the limit point, expand the determinant fully into an ordinary function of the variable first — never try to differentiate or take limits of the determinant "as a whole."

Steps

Step 1: Expand the determinant using cofactor expansion

Treat every entry as an ordinary function of the variable and expand completely, checking each $2\times2$ minor individually — some minors may simplify to zero because their rows/columns are proportional.

Step 2: Simplify and factor out the matching power

Combine the expanded terms and factor out whatever power of the variable matches the limit's denominator (e.g. pull out $t^2$ if the limit divides by $t^2$):

Step 3: Cancel and apply standard limits …

Common Mistakes

Mistake 1: Applying L'Hopital's rule directly to the un-expanded determinant

Why it's wrong: A determinant is not a single differentiable formula that can be blindly differentiated term-by-term as if it were one function; trying this before expanding leads to a tangle of incorrect derivative terms. Correct approach: always fully expand the determinant into an explicit algebraic/trigonometric expression in $t$ first, and only then apply limit techniques to that expression.

Mistake 2: Missing that one of the $2\times2$ minors is identically zero …