Problem 10
Question
Find the limits. $$\lim _{t \rightarrow 0} \frac{t e^{t}}{1-e^{t}}$$
Step-by-Step Solution
Verified Answer
The limit is -1.
1Step 1: Identify the Form
First, evaluate the limit to identify its form. Substitute \( t = 0 \) into \( \frac{t e^t}{1 - e^t} \). We get \( \frac{0 \cdot e^0}{1 - e^0} = \frac{0}{0} \), which is an indeterminate form.
2Step 2: Apply L'Hôpital's Rule
Since the limit is in an indeterminate form \( \frac{0}{0} \), we apply L'Hôpital's Rule. This rule states that if \( \lim_{t \to a} \frac{f(t)}{g(t)} = \frac{0}{0} \) or \( \frac{\infty}{\infty} \), then it can be evaluated as \( \lim_{t \to a} \frac{f'(t)}{g'(t)} \), provided the new limit exists. Differentiate the numerator and the denominator separately.
3Step 3: Differentiate the Numerator
Differentiate the numerator \( t e^t \). Using the product rule \((uv)' = u'v + uv'\), we get: \( \frac{d}{dt}(t e^t) = e^t + t e^t \).
4Step 4: Differentiate the Denominator
Differentiate the denominator \( 1 - e^t \). The derivative of \( 1 \) is \( 0 \) and the derivative of \( -e^t \) is \( -e^t \). Thus, \( \frac{d}{dt}(1 - e^t) = -e^t \).
5Step 5: Calculate the New Limit
Now calculate the new limit. Substitute the derivatives into L'Hôpital's Rule formula: \( \lim_{t \to 0} \frac{e^t + t e^t}{-e^t} \). This simplifies to \( \lim_{t \to 0} \frac{e^t (1 + t)}{-e^t} = \lim_{t \to 0} - (1 + t) \).
6Step 6: Evaluate the Simplified Limit
As \( t \to 0 \), the expression \( -(1 + t) \) becomes \( -(1 + 0) = -1 \). Therefore, the limit is \( -1 \).
Key Concepts
Indeterminate FormsL'Hôpital's RuleDifferentiationProduct Rule
Indeterminate Forms
When tackling calculus problems related to limits, you might encounter the term "indeterminate forms." These forms hint that the traditional methods of direct substitution won't work, and further analysis is often needed. An indeterminate form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) arises when both the numerator and the denominator of a fraction approach zero or infinity simultaneously as the limit variable approaches a certain point.
This is precisely what occurs in our exercise when substituting \( t = 0 \) into \( \frac{t e^t}{1 - e^t} \). It results in \( \frac{0}{0} \), an indeterminate form, signaling that the limit cannot be evaluated in its current state. We'll need a more advanced technique to solve this.
Indeterminate forms don't provide enough information to conclude a limit's value without applying methods like factoring, rationalizing, or rules such as L'Hôpital's Rule.
This is precisely what occurs in our exercise when substituting \( t = 0 \) into \( \frac{t e^t}{1 - e^t} \). It results in \( \frac{0}{0} \), an indeterminate form, signaling that the limit cannot be evaluated in its current state. We'll need a more advanced technique to solve this.
Indeterminate forms don't provide enough information to conclude a limit's value without applying methods like factoring, rationalizing, or rules such as L'Hôpital's Rule.
L'Hôpital's Rule
L'Hôpital's Rule is a fantastic tool for resolving limits that feature indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Named after the French mathematician Guillaume de l'Hôpital, this rule allows us to differentiate the numerator and the denominator separately, and then calculate the limit again on the newly formed fraction.
The critical step is that L'Hôpital's Rule can only be applied when certain conditions are met: both the numerator and denominator should be differentiable, and they should form an indeterminate type initially. In our example, since \( \frac{t e^t}{1 - e^t} \) yields \( \frac{0}{0} \), we have a green light to differentiate both parts.
Upon applying L'Hôpital's Rule to our problem, we simplify the computation and find the actual limit by replacing the expression with its derivatives and checking the new limit.
The critical step is that L'Hôpital's Rule can only be applied when certain conditions are met: both the numerator and denominator should be differentiable, and they should form an indeterminate type initially. In our example, since \( \frac{t e^t}{1 - e^t} \) yields \( \frac{0}{0} \), we have a green light to differentiate both parts.
Upon applying L'Hôpital's Rule to our problem, we simplify the computation and find the actual limit by replacing the expression with its derivatives and checking the new limit.
Differentiation
Differentiation is at the heart of calculus. It involves finding the derivative of a function, which is the rate at which a function changes at a given point. Differentiating functions allows us to understand how output values change concerning changes in input values.
In the solution to the limit problem, we need to differentiate the numerator, \( t e^t \), and the denominator, \( 1 - e^t \). Differentiation requires identifying each component of a function and applying rules such as the product rule, which helps break down more complex expressions.
By differentiating, you transform the problem into a simpler one, making it accessible to calculate the necessary limits and thus solve for the original problem.
In the solution to the limit problem, we need to differentiate the numerator, \( t e^t \), and the denominator, \( 1 - e^t \). Differentiation requires identifying each component of a function and applying rules such as the product rule, which helps break down more complex expressions.
By differentiating, you transform the problem into a simpler one, making it accessible to calculate the necessary limits and thus solve for the original problem.
Product Rule
The product rule is an essential tool in differentiation, especially when dealing with products of two functions. The product rule states: if \( u \) and \( v \) are both differentiable functions of \( t \), then the derivative of their product \( uv \) is \( u'v + uv' \). Thus, to differentiate a product, you differentiate one function and keep the other unchanged, then switch roles and add the two results together.
In the exercise at hand, the numerator \( t e^t \) is a product of two functions: \( t \) and \( e^t \). Using the product rule, the differentiation results in \( e^t + t e^t \). This gives us a simplified expression of the numerator to work with after applying L'Hôpital's Rule.
This rule is crucial because it aids in breaking down complex derivatives into manageable components, enabling a straightforward solution to problems that might otherwise appear intricate.
In the exercise at hand, the numerator \( t e^t \) is a product of two functions: \( t \) and \( e^t \). Using the product rule, the differentiation results in \( e^t + t e^t \). This gives us a simplified expression of the numerator to work with after applying L'Hôpital's Rule.
This rule is crucial because it aids in breaking down complex derivatives into manageable components, enabling a straightforward solution to problems that might otherwise appear intricate.
Other exercises in this chapter
Problem 9
Find \(d y / d x\). $$y=\ln x^{2}$$
View solution Problem 9
Find \(d y / d x\) by implicit differentiation. $$\sin \left(x^{2} y^{2}\right)=x$$
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Suppose that \(z=x^{3} y^{2},\) where both \(x\) and \(y\) are changing with time. At a certain instant when \(x=1\) and \(y=2, x\) is decreasing at the rate of
View solution Problem 10
Confirm that the stated formula is the local linear approximation at \(x_{0}=0\). $$\ln (1+x) \approx x$$
View solution