In this post, we’ll learn a step-by-step approach to solve the following limit problem:
\[ \lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x} \]If we substitute x=0 directly, we encounter a divide-by-zero problem. To solve this, we’ll use a technique called rationalization to ensure the denominator is not zero. Let’s dive into the solution!
Solution:
\( \displaystyle \lim_{x \to 0} \frac{\sqrt{1 + x} – 1}{x} \\ \displaystyle = \lim_{x \to 0} \frac{(\sqrt{1 + x} – 1)(\sqrt{1 + x} + 1)}{x (\sqrt{1 + x} + 1)} \scriptsize \text{, rationalize the numerator by multiplying the numerator and denominator } \\ \scriptsize \text{by the conjugate } \sqrt{1 + x} + 1. \\ \displaystyle = \lim_{x \to 0} \frac{(1 + x) – 1}{x (\sqrt{1 + x} + 1)} \small \text{ , using } (a – b)(a + b) = a^2 – b^2. \\ \displaystyle = \lim_{x \to 0} \frac{x}{x (\sqrt{1 + x} + 1)} \small \text{, cancel } x \text{ since } x \neq 0. \\ \displaystyle = \lim_{x \to 0} \frac{1}{\sqrt{1 + x} + 1} \\ \displaystyle = \frac{1}{\sqrt{1 + 0} + 1} \\ \displaystyle = \frac{1}{1 + 1} \\ \displaystyle = \frac{1}{2} \)Please let me know in the comments if you find any error in this solution.