Evaluating $\lim_{x\to 1} \dfrac{\sin(3x^2 - 5x + 2)}{x^2+x-2}$

Question

How to evaluate the following limit without using the L'Hôspital rule? $$ \lim_{x\to 1} \dfrac{\sin(3x^2 - 5x + 2)}{x^2+x-2} $$

Answer 1

using the fact that $\lim _{ x\rightarrow 0 }{ \frac { \sin { x } }{ x } } =1\\ $ we get $$\lim _{ x\rightarrow 1 }{ \frac { \sin { \left( 3{ x }^{ 2 }-5x+2 \right) } }{ { x }^{ 2 }+x-2 } } =\lim _{ x\rightarrow 1 }{ \frac { \sin { \left( 3{ x }^{ 2 }-5x+2 \right) } }{ 3{ x }^{ 2 }-5x+2 } \frac { 3{ x }^{ 2 }-5x+2 }{ { x }^{ 2 }+x-2 } } =\\ =\lim _{ x\rightarrow 1 }{ \frac { \left( x-1 \right) \left( 3x-2 \right) }{ \left( x+2 \right) \left( x-1 \right) } } =\frac { 1 }{ 3 } $$

Answer 2

Hint: $$ \frac{\sin(3x^2 - 5x + 2)}{x^2+x-2} = \frac{\sin(3x^2 - 5x + 2)}{3x^2 - 5x + 2}\cdot \frac{3x^2 - 5x + 2}{x^2+x-2}. $$

Answer 3

$$\lim_{x\to1} \frac{\sin(3x^2 - 5x + 2)}{x^2+x-2}=\lim_{x\to1} \frac{\sin(x-1)(3x-2)}{(x-1)(3x-2)}\times\dfrac{(3x-2)}{(x+2)}=\dfrac13$$