Question

Let \(f(x)=\left(\sqrt{x+c^{2}}-c\right) / x, c>0 .\) What is the domain of \(f ?\) How can you define \(f\) at \(x=0\) in order for \(f\) to be continuous there?

Short answer

The domain of \(f(x)\) is \(-c^2 < x < 0\) or \(x > 0\). The function \(f(x)\) can be defined at \(x = 0\) as \(f(0)=\frac{1}{2c}\) to make the function continuous there.

Step-by-step solution

Find the domain of the function

The function \(f(x)\) is defined only for those values of \(x\) for which \(\sqrt{x+c^{2}}\) is defined and \(x≠0\) (as \(x\) is in the denominator). As a square root function is defined only for non-negative values, \(x+c^{2}\) should be greater than or equal to 0. Solving this inequality, we get \(x ≥ -c^2 \). Since \(x≠0\), we eliminate 0 from the domain. Therefore, the domain of \(f(x)\) is \(-c^2 < x < 0 \) or \(x > 0\).

Find the limit as x approaches 0 from the positive side

To ensure that the function is continuous at \(x=0\), the limit as \(x\) approaches 0 from the positive side should exist. We calculate the limit using: \[\lim_{{x \to 0}^{+}} f(x) = \lim_{{x \to 0}^{+}} \frac{\sqrt{x+c^{2}}-c}{x}\]. We can simplify the limit using rationalization, which gives us \[\lim_{{x \to 0}^{+}} \frac{x}{x(\sqrt{x+c^2}+c)}\]. After simplification, the limit formula comes down to \[\frac{1}{\sqrt{x+c^2}+c}\]. Substituting \(x = 0\) in the equation, we get \(\frac{1}{2c}\).

Find the limit as x approaches 0 from the negative side

Similarly, we calculate the limit as \(x\) approaches 0 from the negative side using the equation \[\lim_{{x \to 0}^{-}} f(x) = \lim_{{x \to 0}^{-}} \frac{\sqrt{x+c^{2}}-c}{x}\]. Similar to the previous step, the limit as \(x\) approaches 0 from negative side also gives us \(\frac{1}{2c}\).

Evaluating the continuity at x=0

Since the limit exists and they are equal from both the positive and negative side as \(x\) approaches 0, \(f(x)\) is continuous at \(x=0\) if we define \(f(0)=\frac{1}{2c}\)