Find the limit as x approaches 1 of (x^2 - 1) tan(2x - 2) / sin^2(x - 1).

Step-by-step solution

1. 1

   Confirm the indeterminate form. At $x = 1$: $x^2-1 = 0$, $\tan(2x-2) = \tan 0 = 0$ and $\sin^2(x-1) = 0$, so the expression is $\dfrac{0 \cdot 0}{0}$ — genuinely indeterminate, and direct substitution is not allowed.

2. 2

   Substitute $u = x-1$ so everything is centred at zero. Then $x = u+1$, $x^2 - 1 = (x-1)(x+1) = u(u+2)$, $2x-2 = 2u$, and $x-1 = u$, with $u \to 0$:

   $\lim_{u\to 0}\frac{u(u+2)\tan(2u)}{\sin^2 u}$

   This substitution is the whole trick: it exposes the standard small-angle limits.

3. 3

   Rewrite using the two standard limits. Split the expression so each trig factor appears over its own argument:

   $\frac{u(u+2)\tan(2u)}{\sin^2 u} = (u+2)\cdot\underbrace{\frac{\tan(2u)}{2u}}_{\to\,1}\cdot 2 \cdot \underbrace{\left(\frac{u}{\sin u}\right)^{2}}_{\to\,1}$

   Here $u \cdot \dfrac{2u}{u^2} = \dfrac{2}{1}$ accounts for the factor $2$, using $\lim_{t\to0}\tfrac{\tan t}{t} = 1$ and $\lim_{u\to0}\tfrac{\sin u}{u} = 1$.

4. 4

   Take the limit factor by factor.

   $\lim_{u\to 0}(u+2) \cdot 1 \cdot 2 \cdot 1 = 2 \cdot 2 = 4$

5. 5

   Cross-check with leading-order expansions. For small $u$, $\tan(2u) \approx 2u$ and $\sin^2 u \approx u^2$, so the whole quotient behaves like $\dfrac{u(u+2)(2u)}{u^2} = 2(u+2) \to 4$ ✓ — the same answer by a shorter route.

6. 6

   Verify numerically. At $u = 0.01$ (i.e. $x = 1.01$): numerator $= 0.01 \times 2.01 \times \tan(0.02) = 0.01 \times 2.01 \times 0.0200027 = 0.000402053$, denominator $= \sin^2(0.01) = 0.0000999967$, quotient $= 4.02067$. At $u = 0.001$ the quotient is $4.002007$ ✓, closing on $4$, and a symbolic limit returns exactly $4$.