Question

In Exercises 69-71, find the limit. Give a convincing argument that the value is correct. $$\lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\ln x}$$

Short answer

The limit as \(x\) approaches infinity of \(\frac{\ln (x+1)}{\ln x}\) is 1.

Step-by-step solution

Identify the Form of the Expression

The given expression is \(\lim _{x \rightarrow \infty} \frac{\ln (x+1)}{\ln x}\). As \(x\) approaches infinity, both the numerator and the denominator approach infinity, that is, the expression is in the form ∞/∞, which is an indeterminate form.

Apply L'Hopital's Rule

According to L'Hopital's Rule, if a limit is in the indeterminate form 0/0 or ∞/∞, it can be evaluated by differentiating the numerator and the denominator. So, let's find the derivatives of the numerator and the denominator. The derivative of \(\ln (x+1)\) with respect to \(x\) is \(\frac{1}{x+1}\) and the derivative of \(\ln x\) with respect to \(x\) is \(\frac{1}{x}\). So the expression becomes \(\lim _{x \rightarrow \infty} \frac{\frac{1}{x+1}}{\frac{1}{x}} = \lim _{x \rightarrow \infty} \frac{x}{x+1}\).

Simplify the Expression and Evaluate the Limit

The expression \(\lim _{x \rightarrow \infty} \frac{x}{x+1}\) can be simplified by dividing both the numerator and the denominator by \(x\) to obtain \(\lim _{x \rightarrow \infty} \frac{1}{1 + \frac{1}{x}}\). As \(x\) approaches infinity, \(\frac{1}{x}\) approaches 0, and therefore, the expression simplifies to \(\lim _{x \rightarrow \infty} \frac{1}{1 + 0}\), which equals 1.