Question

101.

$$ \\lim _ { x \ ightarrow 1 / 2 } \\frac { 2 x ^ { 2 } + 3 x - 2 } { 2 x - 1 } $$

Explanation:

Step1: Factor the numerator

Factor \(2x^{2}+3x - 2\). We use the formula \(ax^{2}+bx + c=a(x - x_1)(x - x_2)\), where \(x_{1,2}=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\). For \(a = 2\), \(b = 3\), \(c=-2\), \(x=\frac{-3\pm\sqrt{9+16}}{4}=\frac{-3\pm5}{4}\). So \(x_1=\frac{1}{2}\), \(x_2=-2\), and \(2x^{2}+3x - 2=(2x - 1)(x + 2)\)

Step2: Simplify the function

The original limit \(\lim_{x
ightarrow\frac{1}{2}}\frac{2x^{2}+3x - 2}{2x - 1}=\lim_{x
ightarrow\frac{1}{2}}\frac{(2x - 1)(x + 2)}{2x - 1}\). When \(x
eq\frac{1}{2}\), \(\frac{(2x - 1)(x + 2)}{2x - 1}=x + 2\)

Step3: Calculate the limit

Now we find \(\lim_{x
ightarrow\frac{1}{2}}(x + 2)\). Substitute \(x=\frac{1}{2}\) into \(x + 2\), we get \(\frac{1}{2}+2=\frac{1 + 4}{2}=\frac{5}{2}\)

Answer:

\(\frac{5}{2}\)