Subtitusi $x=1$ menghasilkan nilai tak tentu $\frac{0}{0}$

Ingat bahwa

$\lim\limits_{x\rightarrow0}\ \frac{ax}{\sin bx}=\frac{a}{b}$

$\lim\limits_{x\rightarrow0}\ \frac{\tan ax}{\sin bx}=\frac{a}{b}$

Bentuk $\left(x^2-1\right)$ dapat difaktorkan menjadi $x^2-1=\left(x-1\right)\left(x+1\right)$ , maka

$\lim\limits_{x\rightarrow1}\ \frac{\left(x^2-1\right)\tan\left(6x-6\right)}{\sin^2\left(x-1\right)}=\lim\limits_{x\rightarrow1}\ \frac{\left(x-1\right)\left(x+1\right)\tan6\left(x-1\right)}{\sin^2\left(x-1\right)}$

$=\lim\limits_{x\rightarrow1}\ \left(\frac{\left(x-1\right)}{\sin\left(x-1\right)}\cdot\left(x+1\right)\cdot\frac{\tan6\left(x-1\right)}{\sin\left(x-1\right)}\right)$

$=\lim\limits_{x\rightarrow1}\ \frac{\left(x-1\right)}{\sin\left(x-1\right)}\cdot\lim\limits_{x\rightarrow1}\ \left(x+1\right)\cdot\lim\limits_{x\rightarrow1}\ \frac{\tan6\left(x-1\right)}{\sin\left(x-1\right)}$

$=1\cdot\lim\limits_{x\rightarrow1}\ \left(x+1\right)\cdot6$

$=1\cdot\left(1+1\right)\cdot6$

$=12$

Jadi, nilai dari $\lim\limits_{x\rightarrow1}\ \frac{\left(x^2-1\right)\tan\left(6x-6\right)}{\sin^2\left(x-1\right)}=12$