Chapter 2: Problem 80
Evaluate the following limits. \(\lim _{x \rightarrow 0} \frac{x}{\sqrt{c x+1}-1},\) where \(c\) is a nonzero constant
Short Answer
Expert verified
Answer: The value of the given limit is \(\frac{2}{c}\), where \(c\) is a nonzero constant.
Step by step solution
01
Identify the indeterminate form
First, let's see what we get if we substitute x = 0 directly in the expression:
\(\frac{0}{\sqrt{c \cdot 0 + 1} - 1} = \frac{0}{\sqrt{1} - 1} = \frac{0}{0}\)
This is the indeterminate form of the limit, meaning we have to manipulate the expression to find the actual limit.
02
Rationalize the denominator
To find the actual limit, we will need to rationalize the denominator of the expression. Remember, when we rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator.
The conjugate of the denominator \(\sqrt{cx+1} - 1\) is \(\sqrt{cx+1} + 1\).
So, we get:
\(\frac{x}{\sqrt{cx+1}-1}\cdot\frac{\sqrt{cx+1}+1}{\sqrt{cx+1}+1} = \frac{x(\sqrt{cx+1}+1)}{cx+1-1} = \frac{x(\sqrt{cx+1}+1)}{cx}\)
03
Simplify the expression and evaluate the limit
From step 2, we have:
\(\frac{x(\sqrt{cx+1}+1)}{cx}\)
Now, we can simplify by cancelling out the \(x\) from both the numerator and denominator:
\(\frac{\sqrt{cx+1}+1}{c}\)
Now, it is safe to directly substitute the value x = 0 to find the limit:
\(\lim_{x \rightarrow 0} \frac{\sqrt{c x+1}+1}{c} = \frac{\sqrt{c \cdot 0 + 1} + 1}{c} = \frac{\sqrt{1}+1}{c} = \frac{2}{c}\)
So, the limit \(\lim_{x \rightarrow 0} \frac{x}{\sqrt{c x+1}-1}\) is \(\frac{2}{c}\), where \(c\) is a nonzero constant.
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Indeterminate forms
When evaluating limits, you might encounter expressions that seem to yield an undefined form, such as \( \frac{0}{0} \). These are known as indeterminate forms. They tell us that the expression doesn't immediately provide a clear result, and further work is needed. Encountering this form means we can't simply substitute the value into the function to find the limit. Instead, we need clever algebraic manipulations or other limit theorems. This step is crucial as it highlights the need for methods like rationalization or conjugate multiplication to explore the true behavior of the function as it approaches the limit.
Rationalization
Rationalization is a technique used to eliminate irrational expressions in the denominator (or numerator) of a fraction. In limit evaluation, particularly when facing indeterminate forms, rationalization often simplifies the process. The basic idea is to multiply the numerator and denominator of the fraction by a suitable expression that removes the square root or other irrational parts. This step is essential to transform the limit expression into a solvable format. In our exercise, rationalization is needed to handle the square root in the denominator. This technique allows us to rewrite the expression in a form where x can be factored out, making it easier to evaluate the limit.
Conjugate multiplication
Conjugate multiplication is a specific rationalization method used to simplify the expression involving square roots. The conjugate of a binomial containing a square root, such as \( \sqrt{a} - b \), is \( \sqrt{a} + b \). When you multiply these, it eliminates the square root because it results in the difference of squares: \((\sqrt{a} - b)(\sqrt{a} + b) = a - b^2 \). This technique is handy in limit problems like our example where the denominator contains a square root. By multiplying with the conjugate, the expression simplifies to a form where direct substitution becomes possible. Using conjugate multiplication is often a strategic move to resolve indeterminate forms in calculus.
Simplification of expressions
Simplification of expressions is a critical step in limit evaluation, particularly after rationalization and conjugate multiplication. After these processes, expressions often simplify to forms where direct substitution is valid. In this context, simplification involves canceling common factors or expressions, making the limit easier to evaluate. For instance, in our exercise, multiplying by the conjugate left us with the expression \( \frac{x(\sqrt{cx+1}+1)}{cx} \). Here, x is a common factor that can be canceled from both the numerator and denominator. Simplifying in this way exposes the limit expression's true form without the initial indeterminate behavior, allowing us to straightforwardly evaluate the limit at the specified point.