Question

Let \(a\) and \(b\) be positive real numbers. Evaluate \(\lim _{x \rightarrow \infty}(a x-\sqrt{a^{2} x^{2}-b x})\) in terms of \(a\) and \(b\)

Short answer

Question: Evaluate the limit of the expression \((ax-\sqrt{a^2x^2-bx})\) as \(x\) approaches infinity, where \(a\) and \(b\) are positive real numbers. Answer: The limit is \(\frac{b}{2a}\).

Step-by-step solution

Write the given expression in a simplified form

We can start by factoring out \(x\) from the expression inside the square root: \((ax - \sqrt{a^2x^2-bx}) = (ax - \sqrt{x^2(a^2-\frac{b}{x})})\)

Use a common technique (multiplying by conjugate)

One of the common techniques to simplify expressions like this is to multiply and divide the expression by its conjugate. So we have: \(\lim_{x\to\infty} \frac{(ax - \sqrt{x^2(a^2-\frac{b}{x})})(ax + \sqrt{x^2(a^2-\frac{b}{x})})}{(ax + \sqrt{x^2(a^2-\frac{b}{x})})}\)

Simplify the numerator

Now we need to simplify the numerator using the difference of squares formula: \((ax - \sqrt{x^2(a^2-\frac{b}{x})})(ax + \sqrt{x^2(a^2-\frac{b}{x})}) = a^2x^2 - x^2(a^2-\frac{b}{x})\) Now we can further simplify the expression: \(a^2x^2 - x^2(a^2-\frac{b}{x}) = a^2x^2 - a^2x^2 + bx\) So we have: \(\lim_{x\to\infty} \frac{bx}{(ax + \sqrt{x^2(a^2-\frac{b}{x})})}\)

Simplify the limit expression by dividing numerator and denominator by x

In order to find the limit as \(x \rightarrow \infty\), we can simplify the expression further by dividing numerator and denominator by x: \(\lim_{x\to\infty} \frac{b}{(a + \sqrt{a^2-\frac{b}{x}})}\)

Evaluate the limit

As x goes to infinity, the term \(\frac{b}{x}\) in the denominator will go to 0, because b is a constant and x is going to infinity. Therefore, we can rewrite the limit expression as: \(\lim_{x\to\infty} \frac{b}{(a + \sqrt{a^2})}\) Now, we evaluate the limit: \(\frac{b}{(a + \sqrt{a^2})} = \frac{b}{(a + a)} = \frac{b}{2a}\) The evaluated limit of the given expression is \(\frac{b}{2a}\) in terms of \(a\) and \(b\).