Problem 94
Question
Let \(f(x)=g^{\prime}(x) \frac{e^{1 / x}-e^{-1 / x}}{e^{1 / x}+e^{-1 / x}}\), where \(g^{\prime}\) is the derivative of \(g\) and is a continuous function, then \(\lim _{x \rightarrow 0} f(x)\) exists if (A) \(g(x)\) is a polynomial (B) \(g(x)=x\) (C) \(g(x)=x^{2}\) (D) \(g(x)=x^{3} h(x)\), where \(h(x)\) is a polynomial
Step-by-Step Solution
Verified Answer
(C) \( g(x) = x^2 \) and possibly (D) \( g(x) = x^3 h(x) \).
1Step 1: Evaluate the hyperbolic tangent function
The expression \( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \) resembles the hyperbolic tangent function \( \tanh\left(\frac{1}{x}\right) \) for large values of \( \frac{1}{x} \). As \( x \to 0 \), \( \frac{1}{x} \to \infty \) or \( -\infty \), making the expression approach either 1 or -1 due to the nature of the hyperbolic tangent.
2Step 2: Determine \( g'(0) \) and L'Hopital's rule
For \( \lim_{x \to 0} f(x) \) to exist, the term \( g'(x) \) must counterbalance or simplify the behavior of the \( \tanh \left( \frac{1}{x} \right) \) enhanced tendencies. Notice \( f(x) = g'(x)tanh(1/x) \). By L'Hôpital's rule, analyze the limit behavior given that \( f(x) \) has \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) form.
3Step 3: Evaluate given functions
Examine each case:- **(A) \( g(x) \) is a polynomial**: As \( g(x) \to 0 \) and \( g(x) = a_0 + a_1x + \cdots \), analyze terms individually.- **(B) \( g(x) = x \)**: \( g'(x) = 1 \). The expression simplifies to \( \lim_{x \to 0} \tanh \left( \frac{1}{x} \right) \), which does not exist.- **(C) \( g(x) = x^2 \)**: \( g'(x) = 2x \). Substitute in the limit, reshaping to \( \lim_{x \to 0} 2x \tanh\left( \frac{1}{x} \right) = 0 \).- **(D) \( g(x) = x^3 h(x) \)**: \( g'(x) \) becomes \( 3x^2h(x) + x^3h'(x) \). Applying \( 3x^2 \) allows specifics to cancel the infinite oscillation, leading to 0 in this limit.
4Step 4: Conclude
Considering all options:- (A) is incorrect due to general polynomials not having guaranteed cancellations.- (B) is incorrect since the single \( x \) does not provide enough damping.- (C) is correct as \( x^2 \) balances the limit.- (D) could also produce a defined limit based on \( h(x) \)'s characteristics, though primarily \( x^3 \) allows general sufficiency.Options (C) and (D) have mathematical agreements to support limit existence at \( x \to 0 \).
Key Concepts
Hyperbolic FunctionsL'Hôpital's RuleDerivatives and Continuity
Hyperbolic Functions
Hyperbolic functions are similar to trigonometric functions but are based on hyperbolas instead of circles. Among these functions, the hyperbolic tangent function, known as \( \tanh(x) \), is quite significant. This function can be defined using exponential functions: \( \tanh(x) = \frac{e^x - e^{-x}}{e^x + e^{-x}} \).
In the exercise, the expression \( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \) closely resembles the form of \( \tanh \left( \frac{1}{x} \right) \). As \( x \to 0 \), the value of \( \frac{1}{x} \) becomes extremely large or extremely negative, making the hyperbolic tangent approach \( 1 \) or \( -1 \). This oscillation becomes crucial when analyzing limits as \( x \to 0 \).
It is important to remember that hyperbolic functions have properties like their trigonometric counterparts, such as periodicity and symmetry. These properties help in evaluating limits and derivatives where oscillations occur, a frequent scenario that students encounter when learning calculus.
In the exercise, the expression \( \frac{e^{1/x} - e^{-1/x}}{e^{1/x} + e^{-1/x}} \) closely resembles the form of \( \tanh \left( \frac{1}{x} \right) \). As \( x \to 0 \), the value of \( \frac{1}{x} \) becomes extremely large or extremely negative, making the hyperbolic tangent approach \( 1 \) or \( -1 \). This oscillation becomes crucial when analyzing limits as \( x \to 0 \).
It is important to remember that hyperbolic functions have properties like their trigonometric counterparts, such as periodicity and symmetry. These properties help in evaluating limits and derivatives where oscillations occur, a frequent scenario that students encounter when learning calculus.
L'Hôpital's Rule
L'Hôpital's Rule is a powerful tool in calculus to find limits that at first seem to be undefined, especially those that are of the forms \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The rule states that if you have a limit \( \lim_{x \to a} \frac{f(x)}{g(x)} \), and both \( f(x) \) and \( g(x) \) approach \( 0 \) or \( \infty \) as \( x \to a \), this limit can often be found to be \( \lim_{x \to a} \frac{f'(x)}{g'(x)} \) provided the derivative limit exists.
In this specific exercise, the function \( f(x) = g'(x) \tanh(1/x) \) might yield an indeterminate form as \( x \to 0 \). Applying L'Hôpital's Rule helps in evaluating whether the oscillations of \( \tanh \left( \frac{1}{x} \right) \) are counterbalanced by \( g'(x) \).
Using L'Hôpital's Rule effectively involves identifying such problematic forms and then strategically differentiating the numerator and the denominator. Once you identify this setup, calculating the resultant derivatives can simplify the limit evaluation.
In this specific exercise, the function \( f(x) = g'(x) \tanh(1/x) \) might yield an indeterminate form as \( x \to 0 \). Applying L'Hôpital's Rule helps in evaluating whether the oscillations of \( \tanh \left( \frac{1}{x} \right) \) are counterbalanced by \( g'(x) \).
Using L'Hôpital's Rule effectively involves identifying such problematic forms and then strategically differentiating the numerator and the denominator. Once you identify this setup, calculating the resultant derivatives can simplify the limit evaluation.
Derivatives and Continuity
Derivatives give us the rate at which a function changes at any point. In the context of limits, understanding the derivative of a function can help predict how functions behave very near to a point of interest, such as when \( x \rightarrow 0 \).
In our limit problem, \( g'(x) \) is crucial. Different choices of \( g(x) \) lead to differing behaviors of \( g'(x) \) at \( x = 0 \). For instance, if \( g(x) = x^2 \), then \( g'(x) = 2x \), which impacts the oscillation of \( \tanh \left( \frac{1}{x} \right) \) towards zero as \( x \to 0 \).
Continuity also plays a key role here. A continuous function does not have abrupt jumps, and its derivative tends to be well-behaved, or undefined only at isolated points. When ensuring limits exist, determining continuity and differentiability at the point of interest becomes very important.
In our limit problem, \( g'(x) \) is crucial. Different choices of \( g(x) \) lead to differing behaviors of \( g'(x) \) at \( x = 0 \). For instance, if \( g(x) = x^2 \), then \( g'(x) = 2x \), which impacts the oscillation of \( \tanh \left( \frac{1}{x} \right) \) towards zero as \( x \to 0 \).
Continuity also plays a key role here. A continuous function does not have abrupt jumps, and its derivative tends to be well-behaved, or undefined only at isolated points. When ensuring limits exist, determining continuity and differentiability at the point of interest becomes very important.
- Linear functions like \( g(x) = x \) have simple derivatives.
- Quadratic functions, such as \( g(x) = x^2 \), have derivatives \( 2x \), often making them smoother when near zero.
- Higher degree functions, as in \( g(x) = x^3 h(x) \), can have more complex behavior owing to higher powers.
Other exercises in this chapter
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