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Chapter 1: Problem 6

Find the limit (if it exists). If it does not exist, explain why. $$ \lim _{x \rightarrow 4^{-}} \frac{\sqrt{x}-2}{x-4} $$

Short Answer

Expert verified
The limit as \(x\) approaches 4 from the left of the function \(\frac{\sqrt{x}-2}{x-4}\) is \(\frac{1}{4}\).

Step by step solution

01

Direct Substitution

If we substitute \(x = 4\) directly into the function \(\frac{\sqrt{x}-2}{x-4}\), it yields an expression \(\frac{0}{0}\) which is an indeterminate form. This indicates that we need to manipulate the function or use l'hopital's rule to solve the limit.
02

Simplify The Function (Optional)

We can simplify the function by multiplying by the conjugate of the numerator: \( \frac{\sqrt{x} - 2}{x-4} \times \frac{\sqrt{x} + 2}{\sqrt{x} + 2} \) = \( \frac{x - 4}{(x-4)(\sqrt{x} + 2)} \). After canceling \(x-4\) from the numerator and the denominator we get \( \frac{1}{\sqrt{x} + 2} \).
03

Re-Evaluate The Limit

Now we re-evaluate the limit by substituting \(x = 4\) into the simplified function: \( \lim _{x \rightarrow 4^{-}} \frac{1}{\sqrt{x} + 2} = \frac{1}{2 + 2} = \frac{1}{4} \). Thus, the limit of the function as \(x\) approaches 4 from the left is \(\frac{1}{4}\).

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