Question

Determine whether the following statements are true and give an explanation or counterexample. a. By l'Hôpital's Rule, \(\lim _{x \rightarrow 2} \frac{x-2}{x^{2}-1}=\lim _{x \rightarrow 2} \frac{1}{2 x}=\frac{1}{4}\) b. \(\lim _{x \rightarrow 0}(x \sin x)=\lim _{x \rightarrow 0} f(x) g(x)=\) \(\lim _{x \rightarrow 0} f^{\prime}(x) \lim _{x \rightarrow 0} g^{\prime}(x)=\left(\lim _{x \rightarrow 0} 1\right)\left(\lim _{x \rightarrow 0} \cos x\right)=1\) c. \(\lim _{x \rightarrow 0^{+}} x^{1 / x}\) is an indeterminate form. d. The number 1 raised to any fixed power is 1. Therefore, because \((1+x) \rightarrow 1\) as \(x \rightarrow 0,(1+x)^{1 / x} \rightarrow 1\) as \(x \rightarrow 0\) e. The functions \(\ln x^{100}\) and \(\ln x\) have comparable growth rates as \(x \rightarrow \infty\) f. The function \(e^{x}\) grows faster than \(2^{x}\) as \(x \rightarrow \infty\).

Short answer

Question: Determine if the following statements are true or false and provide explanations or counterexamples to justify your conclusions. a. L'Hôpital's rule can be applied to the limit as x approaches 2 of (x-2)/(x^2-1), and the limit is 1/4. b. The limit as x approaches 0 of x*sin(x) can be found by finding the limits of the derivatives of the functions x and sin(x). c. The limit as x approaches 0 from the right of x^(1/x) is an indeterminate form. d. The limit as x approaches 0 of (1+x)^(1/x) is 1. e. As x approaches infinity, the growth rates of log(x^100) and log(x) are comparable. f. As x approaches infinity, the function e^x grows faster than the function 2^x. Provide short explanations for the answers: a. True - L'Hôpital's rule can be applied since both the numerator and the denominator tend to 0 when x=2, and the limit is indeed 1/4. b. False - The correct method to find the limit is by using the squeeze theorem, which gives a limit of 0. c. False - The limit of the given expression is not indeterminate; after solving the limit using L'Hôpital's rule, we find that it is equal to 0. d. False - The correct limit of the given expression is e, not 1. e. True - Both ln(x^100) and ln(x) grow linearly as x approaches infinity, making their growth rates comparable. f. True - Since (e/2)^x grows without bound as x approaches infinity, e^x grows faster than 2^x.

Step-by-step solution

a. L'Hôpital's Rule correctness

To verify the statement, first check if the L'Hôpital's rule can be applied. The original limit is: \(\lim _{x \rightarrow 2} \frac{x-2}{x^{2}-1}\) Both the numerator and the denominator are equal to zero when \(x=2\). So, L'Hôpital's rule can be applied: \(\lim _{x \rightarrow 2} \frac{x-2}{x^{2}-1}=\lim _{x \rightarrow 2} \frac{1}{2 x}\) Now, calculate the limit: \(\lim _{x \rightarrow 2} \frac{1}{2x}=\frac{1}{4}\) The statement is true.

b. Limit of product of functions

The given limit is: \(\lim _{x \rightarrow 0}(x \sin{x})\) We cannot find the limit of the product of two functions by just finding the limits of their derivatives. Therefore, this statement is false. To find the correct limit, we can use the squeeze theorem: As \(-1 \leq \sin{x} \leq 1\), we have: \(-x \leq x \sin{x} \leq x\) Next, find the limit as \(x \rightarrow 0\): \(\lim _{x \rightarrow 0} {-x} = 0\) and \(\lim _{x \rightarrow 0} {x} = 0\) Thus, by the squeeze theorem, \(\lim _{x \rightarrow 0}(x \sin{x})=0\).

c. Limit of power form

Let's analyze the given limit: \(\lim _{x \rightarrow 0^{+}} x^{1 / x}\) We cannot make any conclusions about indeterminate forms without analyzing the limit. Here, we'll take the natural logarithm of the expression and use the fact that \(\lim _{x \rightarrow a} e^{{\ln{f(x)}}} = \lim _{x \rightarrow a} f(x)\): \(\lim _{x \rightarrow 0^{+}} x^{1 / x} = \lim _{x \rightarrow 0^{+}} e^{\frac{\ln{x}}{x}}\) Now, apply L'Hôpital's rule for the exponent: \(\lim _{x \rightarrow 0^{+}} \frac{\ln{x}}{x}=\lim _{x \rightarrow 0^{+}} \frac{1/x}{1}=-\infty\) Therefore, \(\lim _{x \rightarrow 0^{+}} x^{1 / x} = 0\). The statement "the limit is an indeterminate form" is false.

d. Limit of exponential form

For this statement, we'll find the correct limit of the expression \((1 + x)^{1 / x}\) as \(x \rightarrow 0\): Using the definition of the exponential function, we have: \(\lim _{x \rightarrow 0} (1+x)^{1 / x} = \lim _{x \rightarrow 0} e^{\frac{\ln{(1 + x)}}{x}}\) Now, apply L'Hôpital's rule for the exponent: \(\lim _{x \rightarrow 0} \frac{\ln{(1 + x)}}{x}=\lim _{x \rightarrow 0} \frac{1/(1+x)}{1} = 1\) Therefore, the correct limit is \(e^1 = e\). The statement "\((1+x)^{1/x} \rightarrow 1\) as \(x \rightarrow 0\)" is false.

e. Comparing growth rates

As \(x \rightarrow \infty\), the growth rates of \(\ln{x^{100}}\) and \(\ln{x}\) can be compared as follows: \(\ln{x^{100}} = 100\ln{x}\) Both \(\ln{x^{100}}\) and \(\ln{x}\) grow linearly as \(x\rightarrow \infty\). Their growth rates are comparable, so the statement is true.

f. Exponential growth comparison

To compare the growth rates of \(e^x\) and \(2^x\), we can compute the following limit: \(\lim _{x \rightarrow \infty} \frac{e^x}{2^x} = \lim _{x \rightarrow \infty} (e/2)^x\) Since \(\frac{e}{2} > 1\), it follows that \((e/2)^x\) will grow without bound as \(x \rightarrow \infty\). Thus, \(e^x\) grows faster than \(2^x\) as \(x \rightarrow \infty\), and the statement is true.

Key concepts

Limits of Functions

The concept of a limit is fundamental in calculus and helps us understand the behavior of functions as they approach particular points or infinity. A limit examines a function’s value as the input (usually denoted as x) gets indefinitely close to some number. A simple articulation of a limit of a function f(x) as x approaches a value a is symbolized as \(\lim_{x \rightarrow a} f(x)\).
The calculation of limits can lead to various outcomes. When the function approaches a specific value, we say the limit exists, neatly leading to an unequivocal number. However, at times, the limit can be undefined or lead to infinity. The concept is used not just for evaluating expressions but also as a core tool for differentiation and integration, addressing continuous functions, and solving problems in physics and engineering.

Indeterminate Forms

In calculus, indeterminate forms occur when evaluating the limit of a function leads to an expression that is not immediately clear or definitive. Common indeterminate forms include 0/0, \(\infty/\infty\), 0 \times \infty, \(\infty - \infty\), 1^\infty, 0^0, and \(\infty^0\). These forms cannot be evaluated as they stand and require further manipulation or analysis to resolve.

L'Hôpital's Rule is a popular method to tackle such indeterminate forms, especially 0/0 or \(\infty/\infty\). It involves differentiating the numerator and denominator separately and then taking the limit again. However, it's essential to apply l'Hôpital's Rule only when the conditions for its use are met—specifically, when the numerator and the denominator both approach zero or infinity as x approaches some value.

Exponential Growth Comparison

Comparing the growth of exponential functions is a typical application in scientific fields to determine which processes dominate over time. When dealing with exponential functions, the base of the exponent plays a critical role in the function's growth rate. For example, the functions e^x and 2^x both represent exponential growth, but they do not grow at the same rate.

The function e^x grows more rapidly because the base e (Euler's number, approximately 2.71828) is larger than 2. This can be formally demonstrated by taking the limit of their ratio, as x approaches infinity \(\lim_{x \rightarrow \infty} \frac{e^x}{2^x}\). After transforming the ratio into \((e/2)^x\), it is evident that since \(e/2 > 1\), the expression will increase indefinitely, confirming that e^x indeed grows faster than 2^x as x approaches infinity.

Squeeze Theorem

The Squeeze Theorem, also called the Sandwich Theorem or the Pinching Theorem, is an invaluable tool in calculus for finding limits of functions that are otherwise difficult to evaluate directly. The theorem posits that if a function f(x) is always ‘squeezed’ between two other functions g(x) and h(x), and if at a certain point a the limits of g(x) and h(x) are equal, f(x) must also approach that same limit as x reaches a.

Formally, if:

• \(g(x) \leq f(x) \leq h(x)\)
• \(\lim_{x \rightarrow a} g(x) = \lim_{x \rightarrow a} h(x) = L\)

then\(\lim_{x \rightarrow a} f(x) = L\).
This principle is frequently used when dealing with trigonometric limits, as seen in the solution for \(\lim_{x \rightarrow 0}(x \sin x)\), where the Squeeze Theorem helped establish the limit as zero.