Question

The value of $\lim _{x \rightarrow 2}\left(\left(\frac{x^{3}-4 x}{x^{3}-8}\right)^{-1}-\left(\frac{x+\sqrt{2 x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1}\right)$ is (A) \(1 / 2\) (B) 2 (C) 1 (D) None of these

Short answer

After analyzing and solving the given limit expression step by step, we found out that the limit is equal to the inverse of \(3\sqrt{2}\), which did not match any of the given options. Therefore, the answer is (D) None of these.

Step-by-step solution

Simplify the first sub-expression

To simplify the first sub-expression inside the limit, we will factor out the numerator and denominator: \(\left(\frac{x^{3}-4x}{x^{3}-8}\right)^{-1} = \left(\frac{x(x^2-4)}{x^3-8}\right)^{-1} = \left(\frac{x(x+2)(x-2)}{(x-2)(x^2+2x+4)}\right)^{-1}\) Now, we can cancel out the common factors from the numerator and denominator: \(\left(\frac{x(x+2)(x-2)}{(x-2)(x^2+2x+4)}\right)^{-1} = \left(\frac{x(x+2)}{x^2+2x+4}\right)^{-1}\)

Simplify the second sub-expression

Now we will simplify the second sub-expression inside the limit by finding the least common denominator (LCD) to combine the two fractions: \(\left(\frac{x+\sqrt{2x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1} = \left(\frac{(x+\sqrt{2x})(\sqrt{x}+\sqrt{2}) - \sqrt{2}(x-2)}{(x-2)(\sqrt{x}+\sqrt{2})}\right)^{-1}\)

Find common denominator and simplify

Now, we have the simplified sub-expressions, and we can find the LCD to combine them: \(\lim_{x \rightarrow 2}\left(\left(\frac{x(x+2)}{x^2+2x+4}\right)^{-1}-\left(\frac{(x+\sqrt{2x})(\sqrt{x}+\sqrt{2}) - \sqrt{2}(x-2)}{(x-2)(\sqrt{x}+\sqrt{2})}\right)^{-1}\right)\) The least common denominator (LCD) is \((x^2+2x+4)(x-2)(\sqrt{x}+\sqrt{2})\). Therefore, we will multiply the numerators and denominators by the appropriate factors to combine the fractions: \(\lim_{x \rightarrow 2} \left(\frac{(x^2+2x+4)(x-2)(\sqrt{x}+\sqrt{2}) - x(x+2)(x-2)(\sqrt{x}+\sqrt{2})}{(x^2+2x+4)(x-2)(\sqrt{x}+\sqrt{2})}\right)^{-1}\) Now simplify the expression inside the limit: \(\lim_{x \rightarrow 2} \left(\frac{(x^2+2x+4)(\sqrt{x}+\sqrt{2})(x-2)}{(x^2+2x+4)(x-2)(\sqrt{x}+\sqrt{2})}\right)^{-1}\) Cancel out the common factors: \(\lim_{x \rightarrow 2} \left(\frac{x^2+2x+4}{\sqrt{x}+\sqrt{2}}\right)^{-1}\) Now, we can evaluate the limit by substituting x = 2: \(\left(\frac{2^2+2(2)+4}{\sqrt{2}+\sqrt{2}}\right)^{-1} = \left(\frac{12}{2\sqrt{2}}\right)^{-1} = \left(\frac{6}{\sqrt{2}}\right)^{-1}\) Now, multiply the top and bottom with \(\sqrt{2}\): \(\left(\frac{6\sqrt{2}}{2}\right)^{-1} = \left(3\sqrt{2}\right)^{-1}\) So, the value of the given limit is the inverse of \(3\sqrt{2}\): \(\lim_{x \rightarrow 2}\left(\left(\frac{x^{3}-4x}{x^{3}-8}\right)^{-1}-\left(\frac{x+\sqrt{2x}}{x-2}-\frac{\sqrt{2}}{\sqrt{x}-\sqrt{2}}\right)^{-1}\right) = \left(3\sqrt{2}\right)^{-1}\) Thus, the answer is (D) None of these.