Question

Evaluate:$\underset{x\to 1}{\lim }\left(\frac{2x^2+x-3}{3x^3-3x^2+2x-2}\right)$

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

The correct option is A

$1$

Explanation for correct option:

Find the value of $\underset{x\to 1}{\lim }\left(\frac{2x^2+x-3}{3x^3-3x^2+2x-2}\right)$

Consider the given Equation as

$$\begin{gathered} I=\underset{x\to 1}{\lim }\left(\frac{2x^2+x-3}{3x^3-3x^2+2x-2}\right) \\ I=\left(\frac{2\times 1^2+1-3}{3\times 1^3-3\times 1^2+2\times 1-2}\right)[\mathrm{Putting}x\to 1] \\ I=\left(\frac{2+1-3}{3-3+2-2}\right) \\ I=\left(\frac{3-3}{3-3+2-2}\right) \\ I=\frac{0}{0}\left[\text{intermideate form}\right] \end{gathered}$$

Factoring the numerator and denominator

$$\begin{gathered} I=\underset{x\to 1}{\lim }\left(\frac{2x^2+x-3}{3x^3-3x^2+2x-2}\right) \\ I=\underset{x\to 1}{\lim }\left(\frac{2x^2+3x-2x-3}{3x^2\left(x-1\right)+2\left(x-1\right)}\right) \\ I=\underset{x\to 1}{\lim }\left(\frac{\left(2x+3\right)\left(x-1\right)}{\left(3x^2+2\right)\left(x-1\right)}\right) \\ I=\underset{x\to 1}{\lim }\left(\frac{\left(2x+3\right)}{\left(3x^2+2\right)}\right) \\ I=\frac{\left(2\times 1+3\right)}{\left(3\times 1^2+2\right)} \\ I=\frac{\left(2+3\right)}{\left(3+2\right)} \\ I=\frac{5}{5} \\ I=1 \end{gathered}$$

Hence, the correct answer is Option A.