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

Question

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

$(x-1)$ is a common factor for both polynomials, but I do not know how this helps because $x \to 1$ so I cannot use the fundamental trigonometric limit (after multiplying both the numerator and denominator by $(3x^2-5x+2)$.

Any hint?

Answer 1

Based on your attempt, let's factor both polynomials: \begin{align*} \begin{cases} 3x^{2} - 5x + 2 = (3x^{2} - 3x) - (2x - 2) = 3x(x - 1) - 2(x-1) = (3x-2)(x-1)\\\\ x^{2} + x - 2 = (x^{2} + 2x) - (x + 2) = x(x+2) - (x+2) = (x-1)(x+2) \end{cases} \end{align*}

Based on the well known result \begin{align*} \lim_{z\to 0;z\neq0}\frac{\sin(z)}{z} = 1 \end{align*} we can rewrite the proposed limit as \begin{align*} \lim_{x\to 1;x\neq 1}\frac{\sin((3x-2)(x-1))}{(x-1)(x+2)} & = \lim_{x\to 1;x\neq 1}\frac{\sin((3x-2)(x-1))(3x-2)}{(x-1)(3x-2)(x+2)}\\\\ & = \lim_{x\to 1;x\neq 1}\frac{\sin((3x-2)(x-1))}{(3x-2)(x-1)}\times\frac{3x-2}{x+2} = \frac{1}{3} \end{align*}

That's because \begin{align*} \lim_{x\to 1;x\neq 1}\frac{\sin((3x-2)(x-1))}{(3x-2)(x-1)} = \lim_{z\to 0;z\neq 0}\frac{\sin(z)}{z} = 1 \end{align*} where $z = (3x-2)(x-1)$.

Hopefully this helps.

Answer 2

Notice that pluggin $f(1)=\frac{0}{0}$, therefore it is an indeterminate form and we can apply L'Hopitals rule

$$\lim_{x\rightarrow 1}{\frac{\sin{(3x^2-5x+2)}}{x^2+x-2}}=L=\lim_{x\rightarrow 1}{\frac{\cos{(3x^2-5x+2)} (6x-5)}{2x+1}}$$

Replacin $x\rightarrow 1$

$$\frac{\cos{(3-5+2)} (1)}{3}=\boxed{\frac{1}{3}}$$

Answer 3

Factorize: $3x^2-5x+2=(x-1)(3x-2)$ & $x^2+x-2=(x-1)(x+2)$ $$\lim_{x \to 1} \frac{\sin(3x^2-5x+2)}{x^2+x-2}$$$$=\lim_{x \to 1} \frac{\sin((x-1)(3x-2))}{(x-1)(x+2)}$$ $$=\lim_{x \to 1} \frac{\sin((x-1)(3x-2))}{(x-1)(3x-2)}\cdot \frac{3x-2}{x+2}$$ $$=1\cdot \frac13$$$$=\color{blue}{\frac13}$$