Question

Evaluate: lim x→∞ {[2x^(2) +3x-2]^(1/2)}-{[2x^(2) -3x+2]^(1/2)}

Answer 1

Answer on Question #69108 – Math – Real Analysis

Question

Evaluate:

$$\lim x \rightarrow \infty \left\{\left[ 2 x ^ {\wedge} (2) + 3 x - 2 \right] ^ {\wedge} (1 / 2) \right\} - \left\{\left[ 2 x ^ {\wedge} (2) - 3 x + 2 \right] ^ {\wedge} (1 / 2) \right\}$$

Solution

We have

$$\begin{array}{l} \lim _ {x \rightarrow \infty} \sqrt {2 x ^ {2} + 3 x - 2} - \sqrt {2 x ^ {2} - 3 x + 2} = \\ = \lim _ {x \rightarrow \infty} \frac {(\sqrt {2 x ^ {2} + 3 x - 2} - \sqrt {2 x ^ {2} - 3 x + 2}) (\sqrt {2 x ^ {2} + 3 x - 2} + \sqrt {2 x ^ {2} - 3 x + 2})}{\sqrt {2 x ^ {2} + 3 x - 2} + \sqrt {2 x ^ {2} - 3 x + 2}} = \end{array}$$

|We use the formula $(a - b)(a + b) = a^2 - b^2|$

$$\begin{array}{l} = \lim _ {x \rightarrow \infty} \frac {2 x ^ {2} + 3 x - 2 - (2 x ^ {2} - 3 x + 2)}{\sqrt {2 x ^ {2} + 3 x - 2} + \sqrt {2 x ^ {2} - 3 x + 2}} = \lim _ {x \rightarrow \infty} \frac {6 x - 4}{\sqrt {2 x ^ {2} + 3 x - 2} + \sqrt {2 x ^ {2} - 3 x + 2}} = \\ = \lim _ {x \rightarrow \infty} \frac {6 ^ {- \frac {4}{x}}}{\sqrt {2 + \frac {3}{x} - \frac {2}{x ^ {2}}} + \sqrt {2 - \frac {3}{x} + \frac {2}{x ^ {2}}}} = \frac {6}{2 \sqrt {2}} = \frac {3}{\sqrt {2}} = \frac {3 \sqrt {2}}{2}. \end{array}$$

Answer: $\frac{3\sqrt{2}}{2}$

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