Problem 36 - College Algebra and Calculus: An Applied Approach

Question

Find the limit. $$ \lim _{x \rightarrow 3} \frac{\sqrt{x+1}}{x-4} $$

Step-by-Step Solution

Answer

The limit as \(x\) approaches 3 of the function \(\frac{\sqrt{x+1}}{x-4}\) is \(\frac{1}{2}\)

Step 1: Transform the expression

We can add and subtract 4 to the denominator to rewrite the expression suitable to apply L'Hospital's rule: \[ \lim _{x \rightarrow 3} \frac{\sqrt{x+1}}{x-4} = \lim _{x \rightarrow 3} \frac{\sqrt{x+1}}{(x-3)-1} = \lim _{x \rightarrow 3} \frac{\sqrt{x+1}}{(x-3) - (\sqrt{x+1}+\sqrt{x+1})} \]

Step 2: Apply L'Hopital's Rule

We continue by using L'Hopital's Rule, which states that the limit of a ratio of two functions, if the limit approaches the form 0/0 or ∞/∞, equals the limit of their derivatives: \[ \lim _{x \rightarrow 3} \frac{\sqrt{x+1}}{(x-3) - (\sqrt{x+1}+\sqrt{x+1})} = \lim _{x \rightarrow 3} \frac{\frac{d}{dx} \sqrt{x+1}}{\frac{d}{dx}(x-3) - \frac{d}{dx}(\sqrt{x+1}+\sqrt{x+1})} \] Then, differentiate using chain rule (for the numerator) and equality (for the denominator): \[ = \lim _{x \rightarrow 3} \frac{\frac{1}{2}(x+1)^{-\frac{1}{2}}}{1- \frac{1}{2}(x+1)^{-\frac{1}{2}}} \]

Step 3: Simplify and Compute the Limit

Now, we simplify the expression and compute the limit as x approaches 3: \[ = \lim _{x \rightarrow 3} \frac{\frac{1}{2\sqrt{x+1}}}{1-\frac{1}{2\sqrt{x+1}}} = \lim _{x \rightarrow 3} \frac{1}{2(\sqrt{x+1} - 1)} \] Finally, substitute \(x=3\) into the expression to get the limit: \[ = \frac{1}{2(\sqrt{3+1} - 1)} = \frac{1}{2(2-1)} = \frac{1}{2} \]