Problem 30
Question
If \(\lim _{x \rightarrow a} f(x)=1\) and \(\lim _{x \rightarrow a} g(x)=\infty\), then \(\lim _{x \rightarrow a}\\{f(x)\\}^{g(x)}=e^{\lim (f(x)-1) \times g(x)}\) \(\lim _{x \rightarrow 0}\left(\frac{x-1+\cos x}{x}\right)^{\frac{1}{x}}\) is equal to (a) \(e^{1 / 2}\) (b) \(e^{-1 / 2}\) (c) \(e^{1}\) (d) \(\frac{1}{e}\)
Step-by-Step Solution
Verified Answer
The limit is (b) \( e^{-1/2} \).
1Step 1: Identify the Form
First, notice that the limit involves a form similar to \( f(x)^{g(x)} = \left( \frac{x-1+\cos x}{x} \right)^{\frac{1}{x}} \). As \( x \to 0 \), both the base \( \frac{x-1+\cos x}{x} \) and exponent \( \frac{1}{x} \) approach forms that allow the use of the given formula: the base approaches 1 and the exponent approaches infinity.
2Step 2: Determine the Limits of the Functions
Evaluate the limits separately. For \( \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{x-1+\cos x}{x} \), apply L'Hospital's Rule by finding the derivative of the numerator and denominator: the derivative of the numerator \( x-1+\cos x \) is \( 1 - \sin x \), and the denominator is 1. The limit becomes \( \lim_{x \to 0} (1 - \sin x) = 1 \).For \( \lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{1}{x} = \infty \).
3Step 3: Apply the Exponential Limit Formula
Since both conditions satisfy \( f(x) \to 1 \) and \( g(x) \to \infty \), use the given exponential limit formula: \[ \lim_{x \to 0} \left( \frac{x - 1 + \cos x}{x} \right)^{\frac{1}{x}} = e^{\lim_{x \to 0} (f(x) - 1) \times g(x)} \].
4Step 4: Compute \((f(x) - 1) \times g(x)\)
Substitute \( f(x) - 1 = \frac{x - 1 + \cos x}{x} - 1 = -1 + \frac{\cos x}{x} \) and \( g(x) = \frac{1}{x} \) into the expression, giving: \[ (f(x) - 1) \times g(x) = \frac{-1 + \frac{\cos x}{x}}{x} = \frac{-x + \cos x}{x^2} \].
5Step 5: Determine the Resulting Limit
Evaluate \( \lim_{x \to 0} \frac{-x + \cos x}{x^2} \) using L'Hospital's Rule again. Differentiate the numerator and denominator: the numerator \(-x + \cos x\) becomes \(-1 - \sin x\), and the denominator \(x^2\) becomes \(2x\).Applying L'Hospital's Rule yields \( \lim_{x \to 0} \frac{-1 - \sin x}{2x} = \lim_{x \to 0} \frac{-1}{2x} = -\frac{1}{2} \) as \( x \to 0 \).
6Step 6: Solve the Final Expression
Substitute back into the exponential form: \( e^{\lim (f(x) - 1) \times g(x)} = e^{-1/2} \).
Key Concepts
LimitsL'Hôpital's RuleExponential LimitsLimit of a Function
Limits
Limits are a fundamental concept in calculus, allowing us to analyze the behavior of a function as the input approaches a certain value. When dealing with limits, we often want to find out what value a function gets closer to as the input gets infinitely close to a particular point, but not necessarily equal to that point.
Understanding limits helps us tackle situations where a function does not have a clear output at a certain point or where the function might behave unusually. For example, consider the function at the point where it might involve division by zero. Limits help us figure out what result the function approaches without actually hitting the problematic point.
In our exercise, we found limits helpful to analyze the base and exponent of the given expression as they approached forms suitable for further evaluation. The base was shown to approach 1, while the exponent went towards infinity, setting the stage for further analysis using more advanced calculus techniques.
Understanding limits helps us tackle situations where a function does not have a clear output at a certain point or where the function might behave unusually. For example, consider the function at the point where it might involve division by zero. Limits help us figure out what result the function approaches without actually hitting the problematic point.
In our exercise, we found limits helpful to analyze the base and exponent of the given expression as they approached forms suitable for further evaluation. The base was shown to approach 1, while the exponent went towards infinity, setting the stage for further analysis using more advanced calculus techniques.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in differential calculus used to evaluate limits that lead to indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When a function's limit falls into one of these forms, L'Hôpital's Rule allows us to differentiate both the numerator and the denominator until a determinate form is reached.
In our exercise, L'Hôpital's Rule was utilized to find the limits of \( f(x) = \frac{x - 1 + \cos x}{x} \) as \( x \to 0 \). Applying it required taking the derivative of the numerator \( x - 1 + \cos x \), which results in \( 1 - \sin x \), and compared to the derivative of the denominator, which is 1.
This approach helped ascertain that \( f(x) \rightarrow 1 \) and set the groundwork to compute the product \( (f(x) - 1) \times g(x) \) using L'Hôpital's Rule again, ultimately determining the solution to the problem.
In our exercise, L'Hôpital's Rule was utilized to find the limits of \( f(x) = \frac{x - 1 + \cos x}{x} \) as \( x \to 0 \). Applying it required taking the derivative of the numerator \( x - 1 + \cos x \), which results in \( 1 - \sin x \), and compared to the derivative of the denominator, which is 1.
This approach helped ascertain that \( f(x) \rightarrow 1 \) and set the groundwork to compute the product \( (f(x) - 1) \times g(x) \) using L'Hôpital's Rule again, ultimately determining the solution to the problem.
Exponential Limits
Exponential limits often arise in calculus when dealing with expressions involving powers, especially when bases and exponents approach certain critical values. Understanding these limits allows us to handle expressions that appear to grow very fast or approach infinity in more manageable terms.
For our problem, the given expression \( \left( \frac{x-1+\cos x}{x} \right)^{\frac{1}{x}} \) involves both the base approaching 1 and the exponent approaching infinity, forming an expression of type \( 1^{\infty} \). Such forms are known as indeterminate since they give ambiguous results without further evaluation.
The exercise showed that using the knowledge of limits and L'Hôpital's Rule, we can break down such a complex expression into manageable computation steps, eventually expressing the result in the form of an exponential calculation, specifically \( e^{\lim_{x \to 0} (f(x) - 1) \times g(x)} \), leading to the final result.
For our problem, the given expression \( \left( \frac{x-1+\cos x}{x} \right)^{\frac{1}{x}} \) involves both the base approaching 1 and the exponent approaching infinity, forming an expression of type \( 1^{\infty} \). Such forms are known as indeterminate since they give ambiguous results without further evaluation.
The exercise showed that using the knowledge of limits and L'Hôpital's Rule, we can break down such a complex expression into manageable computation steps, eventually expressing the result in the form of an exponential calculation, specifically \( e^{\lim_{x \to 0} (f(x) - 1) \times g(x)} \), leading to the final result.
Limit of a Function
The concept of having a "limit of a function" refers to the value that a function approaches as the input approaches some point. It is a snapshot of the behavior of the function near that point. While limits do not always have to be calculated at a specific value of \( x \), they help provide a better understanding of function behavior around troublesome points.
In our exercise, the limit of \( \frac{x-1+\cos x}{x} \) as \( x \to 0 \) was crucial. Breaking down \( f(x) \) into its limit allowed us to see that it heads towards 1, which aligned with the exponential limit criterion needed. By using these fundamental limit principles, along with differentiations and transformations of the function based on L'Hôpital's Rule, we efficiently resolved the problem, arriving at \( e^{-1/2} \) as the ultimate limit involved in the computation.
Grasping limits and their computation rules enable us to handle dynamic and intricate mathematical expressions systematically and confidently.
In our exercise, the limit of \( \frac{x-1+\cos x}{x} \) as \( x \to 0 \) was crucial. Breaking down \( f(x) \) into its limit allowed us to see that it heads towards 1, which aligned with the exponential limit criterion needed. By using these fundamental limit principles, along with differentiations and transformations of the function based on L'Hôpital's Rule, we efficiently resolved the problem, arriving at \( e^{-1/2} \) as the ultimate limit involved in the computation.
Grasping limits and their computation rules enable us to handle dynamic and intricate mathematical expressions systematically and confidently.
Other exercises in this chapter
Problem 28
Let \(a_{n}=\underbrace{2.99 \ldots 9,}_{n \text { times }} n \in N\) Statement I \(\left[\lim _{n \rightarrow \infty} a_{n}\right]=\lim _{n \rightarrow \infty}
View solution Problem 29
If \(\lim _{x \rightarrow a} f(x)=1\) and \(\lim _{x \rightarrow a} g(x)=\infty\), then \(\lim _{x \rightarrow a}\\{f(x)\\}^{g(x)}=e^{\lim (f(x)-1) \times g(x)}
View solution Problem 32
\(\operatorname{Let} f(x)=\lim _{n \rightarrow-}\left(\cos \sqrt{\frac{x}{n}}\right)^{n}, g(x)=\lim _{n \rightarrow-}(1+x+x \sqrt[n]{e})^{n} .\) Now, consider t
View solution Problem 35
Match the statements of Column I with values of Column II. Column I (A) \(\lim _{x \rightarrow \frac{\pi^{+}}{2}} \tan ^{-1}(\tan x)\) (B) \(\lim _{n \rightarro
View solution