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Magazine Chapter 4: Problem 91
Use analytical methods to evaluate the following limits. $$\lim _{x \rightarrow 1^{+}}\left(\frac{1}{x-1}-\frac{1}{\sqrt{x-1}}\right)$$
Short Answer
Expert verified
Answer: The limit is 0.
Step by step solution
01
Create a common denominator
To combine the two terms in the given function, we need to make their denominators the same. So, we will multiply numerator and denominator of the second term by \(\sqrt{x-1}\). This allows us to combine the terms under a common denominator of \((x-1)\sqrt{x-1}\):
$$\lim _{x \rightarrow 1^{+}}\left(\frac{1}{x-1}-\frac{1}{\sqrt{x-1}}\right) = \lim _{x \rightarrow 1^{+}}\left(\frac{1}{x-1} - \frac{\sqrt{x-1}}{(x-1)\sqrt{x-1}}\right)$$
02
Combine the two terms
Now, we will combine the terms under the common denominator and obtain a single fraction:
$$\lim _{x \rightarrow 1^{+}}\left(\frac{1}{x-1} - \frac{\sqrt{x-1}}{(x-1)\sqrt{x-1}}\right) = \lim _{x \rightarrow 1^{+}}\left(\frac{ \sqrt{x-1} - 1 }{(x-1)\sqrt{x-1}}\right)$$
03
Simplify the numerator
The numerator is in the form \(\sqrt{x-1} - 1\), which approaches 0 when x approaches 1. To avoid the indeterminate form, we can apply the conjugate of the numerator by multiplying both numerator and denominator by \(\sqrt{x-1} + 1\):
$$\lim _{x \rightarrow 1^{+}}\left(\frac{\sqrt{x-1} - 1}{(x-1)\sqrt{x-1}}\right) = \lim _{x \rightarrow 1^{+}}\left(\frac{(\sqrt{x-1} - 1)((\sqrt{x-1} + 1))}{(x-1)\sqrt{x-1}((\sqrt{x-1} + 1))}\right)$$
04
Multiply numerators and denominators
Let's multiply numerators and denominators and simplify the terms:
$$\lim_{x \rightarrow 1^{+}}\left(\frac{(\sqrt{x-1} - 1)((\sqrt{x-1} + 1))}{(x-1)\sqrt{x-1}((\sqrt{x-1} + 1))}\right) = \lim_{x \rightarrow 1^{+}}\left(\frac{(x-1)}{(x-1)\sqrt{x-1}((\sqrt{x-1} + 1))}\right)$$
05
Cancel common factors
We can now cancel the common factor of \((x-1)\) in the numerator and the denominator:
$$\lim_{x \rightarrow 1^{+}}\left(\frac{(x-1)}{(x-1)\sqrt{x-1}((\sqrt{x-1} + 1))}\right) = \lim_{x \rightarrow 1^{+}}\left(\frac{1}{\sqrt{x-1}(\sqrt{x-1} + 1)}\right)$$
06
Evaluate the limit
Now that we have a simplified expression, we can evaluate the limit as x approaches 1 from the positive side:
$$\lim_{x \rightarrow 1^{+}}\left(\frac{1}{\sqrt{x-1}(\sqrt{x-1} + 1)}\right) = \frac{1}{\sqrt{1-1}(\sqrt{1-1} + 1)} = \frac{1}{0(0 + 1)} = \boxed{0}$$
The limit of the given function as x approaches 1 from the positive side is 0.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits in Calculus
In calculus, a limit is a fundamental concept that describes the behavior of a function as its input approaches a certain point, either from the left, the right, or both. It answers the question: What value does the function approach as the input gets arbitrarily close to a number?
Calculating limits is essential in calculus because it deals with the idea of change and can determine the value of a function at points where it is not explicitly defined. For example, if direct substitution results in an undefined expression, limits can often provide a way to understand the behavior of the function near that point through careful analysis.
Calculating limits is essential in calculus because it deals with the idea of change and can determine the value of a function at points where it is not explicitly defined. For example, if direct substitution results in an undefined expression, limits can often provide a way to understand the behavior of the function near that point through careful analysis.
Limit of a Function
The limit of a function is the expected output value that the function approaches as the input value approaches a particular point. Limits can be evaluated from the left (denoted as \( x \rightarrow c^- \)), from the right (denoted as \( x \rightarrow c^+ \)), or from both sides if the function approaches the same value from both directions.
When a function approaches the same value from both sides of a particular point, the function is said to have a limit at that point. However, if the values differ or if the function behaves erratically as it approaches the point, the limit may not exist.
When a function approaches the same value from both sides of a particular point, the function is said to have a limit at that point. However, if the values differ or if the function behaves erratically as it approaches the point, the limit may not exist.
Limits Approaching a Point
When dealing with limits approaching a point, there are several scenarios to consider. If the function can be simplified so that the input value can be directly substituted to find the limit, this is the simplest case. However, in more complex scenarios like the one given in our exercise, direct substitution can lead to indeterminate forms such as \(0/0\) or \(\infty/\infty\).
In such cases, algebraic manipulation, factoring, rationalization, or other methods are used to resolve the indeterminate form and evaluate the limit. Conceptually, this is akin to finding the trend of the function values as they get closer and closer to the point of interest, without actually reaching that point.
In such cases, algebraic manipulation, factoring, rationalization, or other methods are used to resolve the indeterminate form and evaluate the limit. Conceptually, this is akin to finding the trend of the function values as they get closer and closer to the point of interest, without actually reaching that point.
Algebraic Manipulation of Limits
The algebraic manipulation of limits is an essential technique utilized when evaluating limits analytically, especially when faced with forms that are initially indeterminate. Manipulation methods include factoring, expanding, rationalizing, and finding common denominators as shown in the exercise.
For example, when facing a square root in the function, we might rationalize the numerator or denominator to simplify the expression. This often involves multiplying by the conjugate to remove the square root, which was exemplified in the given solution by multiplying both the numerator and denominator by the conjugate \(\sqrt{x-1} + 1\), which allowed for further simplification and cancellation of terms. Ultimately, this process reveals the limit's value by transforming the expression into a form where direct substitution is possible.
For example, when facing a square root in the function, we might rationalize the numerator or denominator to simplify the expression. This often involves multiplying by the conjugate to remove the square root, which was exemplified in the given solution by multiplying both the numerator and denominator by the conjugate \(\sqrt{x-1} + 1\), which allowed for further simplification and cancellation of terms. Ultimately, this process reveals the limit's value by transforming the expression into a form where direct substitution is possible.
