Question

If \(\mathrm{b}<0, \mathrm{~b} \neq-1\) and a is a positive constant then $\lim _{x \rightarrow-\infty} \frac{a+x}{|x|-\sqrt{b^{2} x^{2}+x}}$ equals (A) \(\frac{1}{|b|-1}\) (B) \(\frac{1}{-b-1}\) (C) \(\frac{1}{b-1}\) (D) \(\frac{1}{1-|\mathrm{b}|}\)

Short answer

Based on the step-by-step solution provided, determine the limit of the rational function as \(x\) approaches negative infinity. The limit of the function is \(\frac{1}{1- b}\), and the answer that matches this limit is: (D) \(\frac{1}{1-|b|}\).

Step-by-step solution

Manipulate the function

Let's rewrite the expression under limit. Divide numerator and denominator by \(|x|\): $$\frac{a+x}{|x|-\sqrt{b^{2} x^{2}+x}} = \frac{\frac{a}{|x|} + 1}{1 - \frac{\sqrt{b^{2} x^{2}+x}}{|x|}}$$

Take the limit

Now let's find \(\lim_{x\to-\infty} \frac{a+x}{|x|-\sqrt{b^{2} x^{2}+x}}\): $$\lim_{x\to-\infty} \frac{\frac{a}{|x|} + 1}{1 - \frac{\sqrt{b^{2} x^{2}+x}}{|x|}} = \frac{\lim_{x\to-\infty}(\frac{a}{|x|} + 1)}{\lim_{x\to-\infty}(1 - \frac{\sqrt{b^{2} x^{2}+x}}{|x|})}$$

Calculate the limits separately

We now need to find the limits in the numerator and the denominator separately: $$\lim_{x\to-\infty}\frac{a}{|x|} = 0$$ For \(x \to-\infty\), we have \(\sqrt{b^2x^2+x} \sim -bx\) and \(|x|\sim-x\). Thus, $$\lim_{x \rightarrow-\infty} \frac{\sqrt{b^{2} x^{2}+x}}{|x|} = \lim_{x \rightarrow-\infty} \frac{-bx}{-x} = b$$ Now let's find the limit of the whole expression: $$\lim_{x\to-\infty} \frac{\frac{a}{|x|} + 1}{1 - \frac{\sqrt{b^{2} x^{2}+x}}{|x|}} = \frac{0 + 1}{1 - b} = \frac{1}{1 - b}$$

Match the answer

Comparing the result \(\frac{1}{1 - b}\) with the given options, we find that the correct answer is: (D) \(\frac{1}{1-|b|}\)