Chapter 1: Problem 54
Finding a Limit In Exercises \(47-62,\) find the limit. $$ \lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x-3} $$
Short Answer
Expert verified
The limit of the function as \(x\) approaches 3 is \(\frac{1}{4}\).
Step by step solution
01
Direct Substitution
Start by directly substituting \(x = 3\) into the expression. This results in \(\frac{\sqrt{3+1}-2}{3-3} = \frac{0}{0}\), which is an indeterminate form. Therefore, we can't easily see what the value of the limit is.
02
Multiply by the Conjugate
To work around the indeterminate form, we multiply the expression by the conjugate of the numerator over itself. This technique is called rationalizing the numerator. Therefore, multiply the numerator and denominator by \(\sqrt{x+1}+2\). This looks like, \(\lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x-3} * \frac{\sqrt{x+1}+2}{\sqrt{x+1}+2}\).
03
Simplify the Expression
Simplifying the expression gives:\(\lim _{x \rightarrow 3} \frac{(x+1)-4}{(x-3)(\sqrt{x+1}+2)}\).This simplifies further to:\(\lim _{x \rightarrow 3} \frac{x-3}{(x-3)(\sqrt{x+1}+2)}\).
04
Cancel Common Terms
We can cancel out the \(x-3\) terms in the numerator and denominator leaving us with the following limit expression: \(\lim_{x \rightarrow 3} \frac{1}{\sqrt{x+1}+2}\)
05
Evaluate the limit
Finally, since there's no more indeterminate form, we substitute \(x=3\) into the simplified expression to get \( \frac{1}{\sqrt{3+1}+2} = \frac{1}{4} \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate Forms
In calculus, indeterminate forms present challenges when evaluating limits. They occur when the limit of an expression is not readily apparent, and direct substitution results in an ambiguity, such as \( \frac{0}{0} \) or \( \frac{\text{\infty}}{\text{\infty}} \). In our example, when trying to find \( \lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x-3} \), direct substitution of x = 3 yields \( \frac{0}{0} \) - a classic indeterminate form. This makes the true limit unclear, as various functions approaching such forms can have vastly different limits. To proceed, one must manipulate the expression to eliminate the indeterminate form and allow for proper limit evaluation.
Understanding how to handle indeterminate forms is crucial in calculus, as they frequently appear during the evaluation process. Techniques such as rationalizing the numerator, applying L'Hospital's rule, or finding a common denominator are common ways to resolve these ambiguities and determine the accurate limit of an expression.
Understanding how to handle indeterminate forms is crucial in calculus, as they frequently appear during the evaluation process. Techniques such as rationalizing the numerator, applying L'Hospital's rule, or finding a common denominator are common ways to resolve these ambiguities and determine the accurate limit of an expression.
Rationalizing the Numerator
Rationalizing the numerator is a technique used to resolve indeterminate forms, specifically when dealing with limits involving roots. It involves multiplying the numerator and denominator by a conjugate pair, which is a binomial formed by changing the sign of the second term in a binomial. This method helps in getting rid of radicals from the numerator.
In the given problem, we use the conjugate of \( \sqrt{x+1}-2 \) to transform the original expression \( \frac{\sqrt{x+1}-2}{x-3} \) into a form that allows us to cancel out the indeterminate nature. This multiplication strategically creates a difference of squares in the numerator, leading to simplified terms that ultimately help in revealing the true limit as x tends to a particular value. Rationalizing proves to be a powerful algebraic tool in limit evaluation, offering clarity to otherwise problematic expressions.
In the given problem, we use the conjugate of \( \sqrt{x+1}-2 \) to transform the original expression \( \frac{\sqrt{x+1}-2}{x-3} \) into a form that allows us to cancel out the indeterminate nature. This multiplication strategically creates a difference of squares in the numerator, leading to simplified terms that ultimately help in revealing the true limit as x tends to a particular value. Rationalizing proves to be a powerful algebraic tool in limit evaluation, offering clarity to otherwise problematic expressions.
Limit Evaluation
Limit evaluation is the process of determining the value that a function approaches as the input approaches a specific point. One of the key principles in calculus, the concept of a limit, captures the idea of approximation and serves as a foundation for other concepts like derivatives and integrals.
Once the indeterminate form is resolved, as in Step 4 of our solution by canceling common terms, the expression no longer has variables in a form that yield zero or infinity when substituted directly. It is at this stage we can perform the limit evaluation by substituting the target x-value into the simplified expression. This final step revealed that \( \lim_{x \rightarrow 3} \frac{1}{\sqrt{x+1}+2} = \frac{1}{4} \), providing a concrete value for the limit. In general, the ability to properly evaluate a limit is pivotal for students to understand how functions behave near a given point and is a skill extensively applied across various fields involving calculus.
Once the indeterminate form is resolved, as in Step 4 of our solution by canceling common terms, the expression no longer has variables in a form that yield zero or infinity when substituted directly. It is at this stage we can perform the limit evaluation by substituting the target x-value into the simplified expression. This final step revealed that \( \lim_{x \rightarrow 3} \frac{1}{\sqrt{x+1}+2} = \frac{1}{4} \), providing a concrete value for the limit. In general, the ability to properly evaluate a limit is pivotal for students to understand how functions behave near a given point and is a skill extensively applied across various fields involving calculus.
Limits at a Point
Considering 'limits at a point' in calculus refers to the behavior of a function as its input nears some specific value. They tell us the value a function approaches (but does not necessarily reach) as the input comes arbitrarily close to a given number. Limits can be approached from both the left (\( x \rightarrow c^- \)) and the right (\( x \rightarrow c^+ \)), and for the limit to exist, both must converge to the same value.
The concept of 'limits at a point' is exemplified in our exercise when we're asked to find \( \lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x-3} \). Here, x is approaching the point 3, and our analysis centers around this specific value. Deep understanding of limits at a point is vital as they underpin the continuity of functions - a fundamental concept in calculus that has countless applications including in physics, engineering, and economics.
The concept of 'limits at a point' is exemplified in our exercise when we're asked to find \( \lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x-3} \). Here, x is approaching the point 3, and our analysis centers around this specific value. Deep understanding of limits at a point is vital as they underpin the continuity of functions - a fundamental concept in calculus that has countless applications including in physics, engineering, and economics.
