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Chapter 3: Problem 13

Find the indicated limits. $$\lim _{x \rightarrow 1} \frac{\sqrt{x}-1}{x-1}$$

Short Answer

Expert verified
The value of the limit is \(\frac{1}{2}\).

Step by step solution

01

Identifying the Problem

The original limit, as \(x\) approaches 1, is in the form \(\frac{0}{0}\), which is indeterminate. Therefore, this needs to be reformatted.
02

Rationalizing the Numerator

Rationalize the numerator to reformat the expression. Multiply the fraction by \(\frac{\sqrt{x}+1}{\sqrt{x}+1}\). The expression then becomes \(\lim_{x \rightarrow 1} \frac{x-1}{(x-1)(\sqrt{x}+1)}\) .
03

Simplifying the New Expression

Solve the problem by simplifying the new expression. The \(x-1\) in the numerator and denominator will cancel out, leaving \(\lim_{x \rightarrow 1} \frac{1}{\sqrt{x}+1}\).
04

Substituting the Value

Now we can substitute \(x=1\) into the expression to find the limit and the expression becomes \(\frac{1}{\sqrt{1}+1}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Indeterminate Forms
In calculus, encountering an indeterminate form can sometimes feel like hitting a wall, but it's just a cue for using clever algebraic tricks. The expression is termed 'indeterminate' when both the numerator and the denominator approach values that do not allow for simple division, such as 0/0. In this exercise, the limit \[ \lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} \] presents such a situation. As \(x\) approaches 1, both \(\sqrt{x} - 1\) and \(x - 1\) approach 0, leading to the 0/0 indeterminate form. These expressions require us to employ techniques like rationalization or simplification to resolve the indeterminate form into something manageable.
Rationalizing the Numerator
Rationalizing is a technique often used to simplify expressions involving roots or radicals. In this context, by multiplying the original expression \[ \frac{\sqrt{x} - 1}{x - 1} \] by the conjugate of the numerator, \(\frac{\sqrt{x}+1}{\sqrt{x}+1}\), we transform the original form into something more workable. This multiplication uses the difference of squares identity, turning the numerator into
  • \( (\sqrt{x})^2 - 1^2 = x - 1 \)
Multiplication by the conjugate introduces a handy factor of \(x - 1\), which can potentially be cancelled out later. After rationalizing, our expression becomes \[ \lim_{x \to 1} \frac{x - 1}{(x - 1)(\sqrt{x} + 1)} \]. This step cleverly circumvents the indeterminate form.
Simplifying Expressions
After rationalization, the new expression \( \frac{x - 1}{(x - 1)(\sqrt{x} + 1)} \) can be simplified. Here, the factor \((x-1)\) appears in both the numerator and the denominator. We can cancel this common factor, provided we are not dividing by zero around the point x=1. It simplifies our expression to \[ \lim_{x \to 1} \frac{1}{\sqrt{x}+1} \].With this step, the expression is now easier to handle, having removed the offending \(0/0\) form that makes direct substitution problematic. Simplification is crucial in limits as it often directly leads to the evaluation of the limit at the desired point.
Substitution Method
Once simplification has been applied, and the indeterminate form is resolved, the substitution method allows us to find the value of the limit. For this specific problem, the limit expression reduces to \[ \frac{1}{\sqrt{x} + 1} \]. By substituting x = 1 directly into this simplified expression, \[ \frac{1}{\sqrt{1} + 1} = \frac{1}{1 + 1} = \frac{1}{2}. \] This process confirms the limit as \(\frac{1}{2}\). Substitution is typically the final step in evaluating limits because it directly addresses the value towards which our function is heading as \(x\) approaches a certain point.

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Most popular questions from this chapter

In this exercise, you will play the role of professor and construct a tricky graphing exercise. The first goal is to find a function with local extrema so close together that they're difficult to see. For instance, suppose you want local extrema at \(x=-0.1\) and \(x=0.1 .\) Explain why you could start with \(f^{\prime}(x)=(x-0.1)(x+0.1)=x^{2}-0.01 .\) Look for a function whose derivative is as given. Graph your function to see if the extrema are "hidden." Next, construct a polynomial of degree 4 with two extrema very near \(x=1\) and another near \(x=0\)

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