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

Answer: The limit is 0.

Step-by-step solution

Substitute

To simplify the expression, let's substitute \(u = ax\). So, the expression becomes \(u-\sqrt{u^2-b(u/a)}\). Now, we need to find the limit as \(u\rightarrow\infty\).

Factor out u from the square root

Let's factor out \(u\) from the square root: \(u-\sqrt{u^2-\frac{b}{a}u}\). Now notice that we can rewrite it as \(u\left(1-\sqrt{1-\frac{b}{au}}\right)\).

Apply limit properties

Now we need to find \(\lim_{u\rightarrow\infty}\left(1-\sqrt{1-\frac{b}{au}}\right)\). We can rewrite the expression as the product of two limits: \(\lim_{u\rightarrow\infty} u \cdot \lim_{u\rightarrow\infty}\left(1-\sqrt{1-\frac{b}{au}}\right)\).

Evaluate the second limit

To evaluate the second limit, notice that as \(u\rightarrow\infty\), the term inside the square root approaches 1: \(\lim_{u\rightarrow\infty}\sqrt{1-\frac{b}{au}} = \lim_{u\rightarrow\infty}\sqrt{1-0}=\lim_{u\rightarrow\infty}1\). Thus, the second limit becomes \(\lim_{u\rightarrow\infty}\left(1-\sqrt{1-\frac{b}{au}}\right) = \lim_{u\rightarrow\infty}\left(1-1\right) = 0\).

Evaluate the product of limits

Now we have \(\lim_{u\rightarrow\infty} u \cdot \lim_{u\rightarrow\infty}\left(1-\sqrt{1-\frac{b}{au}}\right) = \lim_{u\rightarrow\infty} u \cdot 0 = 0\). This means that \(\lim_{x\rightarrow\infty}(ax-\sqrt{a^2x^2-bx})=0\) in terms of \(a\) and \(b\).