101. $$ \\lim _ { x \\rightarrow 1 / 2 } \\frac { 2 x ^ { 2 } + 3 x - 2 } { 2 x - 1 } $$
101. $$ \\lim _ { x \\rightarrow 1 / 2 } \\frac { 2 x ^ { 2 } + 3 x - 2 } { 2 x - 1 } $$
Answer
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\rightarrow\frac{1}{2}}\frac{2x^{2}+3x - 2}{2x - 1}=\lim_{x\rightarrow\frac{1}{2}}\frac{(2x - 1)(x + 2)}{2x - 1}). When (x\neq\frac{1}{2}), (\frac{(2x - 1)(x + 2)}{2x - 1}=x + 2)
Step3: Calculate the limit
Now we find (\lim_{x\rightarrow\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})