Problem 9
Question
Use l'Hopital's rule to find the limits in Exercises \(7-50\) . $$ \lim _{t \rightarrow-3} \frac{t^{3}-4 t+15}{t^{2}-t-12} $$
Step-by-Step Solution
Verified Answer
The limit is \(-\frac{23}{7}\).
1Step 1: Identify the Indeterminate Form
First, we need to verify if the given limit is in an indeterminate form by substituting the limit value into the function. For the numerator, \(t^3 - 4t + 15\), substitute \(t = -3\): \((-3)^3 - 4(-3) + 15 = -27 + 12 + 15 = 0\). For the denominator, \(t^2 - t - 12\), substitute \(t = -3\): \((-3)^2 - (-3) - 12 = 9 + 3 - 12 = 0\). Thus, the function \(\frac{t^{3}-4t+15}{t^{2}-t-12}\) is indeed in the indeterminate form \(\frac{0}{0}\).
2Step 2: Apply L'Hopital's Rule
Since the function is in the \(\frac{0}{0}\) form, we can apply L'Hopital's Rule, which states that \(\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\) provided this limit exists. Compute the derivatives: the derivative of the numerator \(f(t) = t^3 - 4t + 15\) is \(f'(t) = 3t^2 - 4\), and the derivative of the denominator \(g(t) = t^2 - t - 12\) is \(g'(t) = 2t - 1\).
3Step 3: Compute the New Limit
Now substitute back \(t = -3\) into the differentiated function: \(\lim_{t \to -3} \frac{3t^2 - 4}{2t - 1}\). Calculate each part: For \(3t^2 - 4\), substitute \(t = -3\): \(3(-3)^2 - 4 = 3(9) - 4 = 27 - 4 = 23\). For \(2t - 1\), substitute \(t = -3\): \(2(-3) - 1 = -6 - 1 = -7\). So, the limit is \(\frac{23}{-7}\).
4Step 4: Conclude the Solution
We have simplified the problem using L'Hopital's Rule and found the limit of the expression as \(t\) approaches \(-3\). The computation gives us the final result of \(\frac{23}{-7}\).
Key Concepts
Indeterminate FormsDerivativesLimits in CalculusRational Functions
Indeterminate Forms
In calculus, indeterminate forms arise when evaluating limits and result in ambiguous expressions such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms do not provide direct answers and require special techniques to resolve. When you substitute a limit into a function and end up with 0 in the numerator and 0 in the denominator, it's a signal that you've encountered an indeterminate form. This is often the case with complicated functions, and it indicates that further work is needed to evaluate the limit accurately. With indeterminate forms, applying l'Hopital's Rule is a common approach to finding limits.
Derivatives
Derivatives measure how a function changes as its input changes. They provide the slope of a function at any given point and are central to the concept of differential calculus. When using l'Hopital's Rule, derivatives play a crucial role. The rule requires you to differentiate both the numerator and the denominator to simplify the original expression. For example, given a function \(f(x)\) in the numerator and \(g(x)\) in the denominator, their derivatives, \(f'(x)\) and \(g'(x)\), are used to find \(\lim_{x \to a} \frac{f(x)}{g(x)}\) by evaluating \(\lim_{x \to a} \frac{f'(x)}{g'(x)}\). The derivatives effectively remove the indeterminate form, allowing for straightforward limit evaluation.
Limits in Calculus
Limits are foundational in calculus, describing the behavior of a function as its arguments approach a specific value. They help us understand how functions behave near points of interest, such as boundaries and discontinuities. Calculating a limit gives insight that direct substitution might not reveal, especially if it results in an indeterminate form. In the context of l'Hopital's Rule, limits allow us to evaluate expressions that initially seem unsimplifiable, like \(\frac{0}{0}\). By taking derivatives and then re-evaluating the limit, we glean meaningful information about the function's behavior near the point of interest.
Rational Functions
Rational functions are quotients of two polynomials, such as \(\frac{t^3 - 4t + 15}{t^2 - t - 12}\). They often present simplified behavior on an interval, but can become complex around points where the denominator is zero. This results in indeterminate forms when computing limits. The process of analyzing and simplifying rational functions involves techniques like factoring, cancellation, and differentiation. These processes help identify valid limits at points where direct computation isn't possible. Applying calculus tools such as the derivative via l'Hopital's Rule assists in understanding the behavior of rational functions in scenarios that initially result in indeterminate forms.
Other exercises in this chapter
Problem 9
True, or false? As \(x \rightarrow \infty\) $$ \begin{array}{ll}{\text { a. }} & {x=o(x)} & {\text { b. } x=o(x+5)} \\\ {\text { c. }} & {x=O(x+5)} & {\text { d
View solution Problem 9
Solve the differential equations in Exercises \(9-22\) $$2 \sqrt{x y} \frac{d y}{d x}=1, \quad x, y>0$$
View solution Problem 9
Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. \begin{equation}y=e^{5-7 x}\end{equation}
View solution Problem 9
In Exercises \(7-38,\) find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$ y=\ln \left(t^{2}\right) $$
View solution