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Engineering Context
MAE:
Mechanical Engineers use L’Hopital’s rule to calculate the efficiency of a system like a gearbox. The efficiency can be modeled by the function below, where output is the function representing the output power, and input is the function representing the input power. For our example, let’s say both are in terms of \( t \):
\[ \frac{\text{Output}}{\text{Input}} \ast 100 \]
To evaluate the efficiency, we would calculate the limit of this function. L’Hopital’s rule would be used if the input and output functions are too complex to directly compute the limit.
ECE:
Electrical Engineers use limits to evaluate functions of current or voltage. Using a limit allows engineers to predict potential problems such as power surges or power outages. Any time limits are being evaluated, there is a chance that the limit will be indeterminate. This is when L’Hopital’s rule must be used. For example, a function like the one below might be used to model a current through a circuit:
\( f(x) = \frac{e^{2x} - 1}{x} \)
Since taking this function’s limit directly leads to the indeterminate form \( \frac{\infty}{\infty} \), the limit calculation would require use of L’Hopital’s rule.
BE:
Biological engineers could use L’Hopital’s Rule when predicting bacteria growth. If a biological engineer wanted to compare the bacteria growth of two cultures, they could approximate the first culture with a function \( f(x) \) and the second culture with a function \( g(x) \). Using a limit, they could evaluate which culture grows faster.
If
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = \infty \]
then we know that \( f(x) \) is growing faster.
If instead
\[ \lim_{x \to \infty} \frac{f(x)}{g(x)} = 0 \]
we know that \( g(x) \) is growing faster.
L’Hopital’s Rule would be used here because the original limit is indeterminate.
CEE:
L’Hopital’s rule is useful to Civil Engineers in understanding the deflection of a bridge. Since bridges often move in a complex combination of waves, L’Hopital’s rule can be used to simplify the equation. Let's say, for example, that a bridge’s deflection can be modeled by:
\[ \frac{\sin(x) - x^2}{\sin(x)} \]
When we take the limit of this function, we can understand the deflection as the load approaches a value:
\[ \lim_{x \to 0} \frac{\sin(x) - x^2}{\sin(x)} = \frac{0}{0} \]
Using L’Hopital’s rule:
\[ \lim_{x \to 0} \frac{\cos(x) - 2x}{\cos(x)} = \frac{1}{1} = 1 \]
We can find the value for the deflection of a bridge.
The Essentials
L’Hopital’s Rule is a tool used to calculate limits using derivatives. Sometimes, when taking a direct limit, the result is in indeterminate form such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \), and it’s difficult to know how the function will behave at the limit. L’Hopital’s rule can be used to simplify such limit equations so that the direct limit will be determinable.
L’Hopital’s rule states that:
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
It is important to note that we are not calculating the derivative using the quotient rule, but rather the derivative of the numerator and the denominator independently.
A Deeper Dive
The proof for L’Hopital’s rule assumes that the functions f, g, f′, and g′ are continuous over an interval containing the point a. We must also state that \( \lim_{x \to a} f(x) = 0 = \lim_{x \to a} g(x) \). Since this is the case, \( f(a) = 0 = g(a) \). Therefore,
\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f(x) - f(a)}{g(x) - g(a)} \]
We can simplify this by the following steps:
- Algebraically rearrange the equation:
\[ = \lim_{x \to a} \frac{\frac{f(x) - f(a)}{x - a}}{\frac{g(x) - g(a)}{x - a}} \]
- Use the quotient rule for limits:
\[ = \frac{\lim_{x \to a} \frac{f(x) - f(a)}{x - a}}{\lim_{x \to a} \frac{g(x) - g(a)}{x - a}} \]
- Apply the limit definition of a derivative:
\[ = \frac{f'(a)}{g'(a)} \]
- We assume the continuity of \( f' \) and \( g' \):
\[ = \frac{\lim_{x \to a} f'(x)}{\lim_{x \to a} g'(x)} \]
- Use the quotient rule for limits:
\[ = \lim_{x \to a} \frac{f'(x)}{g'(x)} \]
Other Indeterminate Forms
There are several other indeterminate forms including \( 0 · \infty \), \( \infty − \infty \), \( 1^\infty \),\( \infty^0 \), and\( 0^0 \). In order to evaluate these usingL’Hopital’s Rule we must rewrite the indeterminate form to be in quotient form such that we arrive at a case that is either\( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
Practice
1) \( \lim_{x \to \infty} \frac{7x^2 + 3}{9x + 1} \)
2) \( \lim_{x \to \infty} \frac{20x}{x^3 + x} \)
3) \( \lim_{x \to \infty} \frac{5x + 4}{9x + 1} \)
Solution:
If we took these limits directly, we would get \(\frac{\infty}{\infty} \), which is an indeterminate form. To solve this problem we use L’Hopital’s rule and take the derivative of the numerator and denominator:
1) \( \lim_{x \to \infty} \frac{7x^2 + 3}{9x + 1} = \lim_{x \to \infty} \frac{14x}{9} = \infty \)
2) \( \lim_{x \to \infty} \frac{20x}{x^3 + x} = \lim_{x \to \infty} \frac{20}{3x^2 + 1} = 0 \)
3) \( \lim_{x \to \infty} \frac{5x + 4}{9x + 1} = \lim_{x \to \infty} \frac{5}{9} = \frac{5}{9} \)
