L'Hôpital's Rule
Graph $y=\dfrac{e^x-1}{\sin x}$ in Desmos and you see a smooth curve with a single hole at $x=0$. Plug in $x=0$ yourself and you get $\frac{0}{0}$, which tells you nothing. How high should that hole be? You can't factor or rationalize your way out of this one. The derivative can.
The Rule: Compare Derivatives Instead of Functions
When a limit has the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, both the top and the bottom are "fighting" to see which one reaches its target faster. The derivative measures exactly that: how fast each function is changing.
Calculate the following limit using L'Hôpital's Rule, first verifying that the necessary indeterminate form is present: $$\lim_{x \to 2} \frac{x^3 - 5x + 2}{x^2 - 4}$$
Calculate the following limit using L'Hôpital's Rule, first verifying that the necessary indeterminate form is present:
$$\lim_{x \to 2} \frac{x^3 - 5x + 2}{x^2 - 4}$$In the study of the propagation of an electromagnetic signal, the following expression appears to calculate the intensity at a critical point. Find the value of the…
In the study of the propagation of an electromagnetic signal, the following expression appears to calculate the intensity at a critical point. Find the value of the limit as the time $t$ approaches $0$:
$$\lim_{t \to 0} \frac{e^{3t} - 1}{\sin(2t)}$$Determine the value of the following limit of a logarithmic function. Remember to differentiate the numerator and the denominator correctly and independently: $$\lim_{x…
Determine the value of the following limit of a logarithmic function. Remember to differentiate the numerator and the denominator correctly and independently:
$$\lim_{x \to 1} \frac{\ln(x)}{x^2 + 2x - 3}$$Reading the Rule on a Graph: Ratio of Slopes
Here is the idea behind the rule in its simplest case. If $f(a)=g(a)=0$, both graphs pass through the same point on the $x$-axis, and near $a$ they look like lines.
Observe the following graph representing two functions $f(x)$ (in white) and $g(x)$ (in dark gray) that are zero at the point $x = a$. Knowing that $f'(a) = 6$ and…
Observe the following graph representing two functions $f(x)$ (in white) and $g(x)$ (in dark gray) that are zero at the point $x = a$. Knowing that $f'(a) = 6$ and $g'(a) = 2$, calculate the value of the limit:
$$\lim_{x \to a} \frac{f(x)}{g(x)}$$Find the value of the parameter $k$ so that the following limit is equal to $4$: $$\lim_{x \to 0} \frac{\tan(kx)}{x} = 4$$
Find the value of the parameter $k$ so that the following limit is equal to $4$:
$$\lim_{x \to 0} \frac{\tan(kx)}{x} = 4$$Applying the Rule More Than Once
Sometimes one derivative isn't enough: the new quotient is again $\frac00$. That is fine. Check the form, then differentiate again.
It is known that the concentration of a chemical reactant in a mixture follows the function $C(t)$. Calculate the limiting concentration as the time $t$ approaches $0$…
It is known that the concentration of a chemical reactant in a mixture follows the function $C(t)$. Calculate the limiting concentration as the time $t$ approaches $0$ from the right:
$$\lim_{t \to 0^+} \frac{1 - \cos(t)}{t^2}$$Given the functions $f(x) = e^x - x - 1$ and $g(x) = x^2$, find the limit of their quotient as $x$ approaches $0$: $$\lim_{x \to 0} \frac{f(x)}{g(x)}$$ Note: You may…
Given the functions $f(x) = e^x - x - 1$ and $g(x) = x^2$, find the limit of their quotient as $x$ approaches $0$:
$$\lim_{x \to 0} \frac{f(x)}{g(x)}$$Note: You may need to apply L'Hôpital's Rule more than once.
In the study of the propagation of a Wi-Fi signal in an environment with interference, the loss of intensity has been modeled by the following function of the distance…
In the study of the propagation of a Wi-Fi signal in an environment with interference, the loss of intensity has been modeled by the following function of the distance $x$ (in feet):
$$f(x) = \frac{e^{2x} - e^{-2x} - 4x}{x - \sin(x)}$$Calculate the limit of this function as the distance $x$ approaches $0$, that is, $\lim_{x \to 0} f(x)$, to determine the behavior of the signal at the point of emission.
The $\infty/\infty$ Form and Limits at Infinity
The rule works the same way when the top and bottom both grow without bound, and $a$ may be $+\infty$ or $-\infty$.
For a rational function such as $\frac{5x^2+1}{2x^2+3}$ you can compare degrees, but the rule gives the same answer and it also justifies it.
Calculate the limit of the following rational function at infinity. Although you could solve it by comparing degrees, use L'Hôpital's Rule to prove the result: $$\lim_{x…
Calculate the limit of the following rational function at infinity. Although you could solve it by comparing degrees, use L'Hôpital's Rule to prove the result:
$$\lim_{x \to +\infty} \frac{5x^2 - 3x + 1}{2x^2 + 7}$$It is desired to study the convergence of an artificial intelligence algorithm. The error rate as a function of the number of iterations $x$ is defined as: $$E(x) =…
It is desired to study the convergence of an artificial intelligence algorithm. The error rate as a function of the number of iterations $x$ is defined as:
$$E(x) = \frac{\ln(x^4 + 1)}{x}$$Prove, using L'Hôpital's Rule, that as the number of iterations approaches infinity ($x \to \infty$), the error rate approaches $0$. What does this tell us about the long-term efficiency of the algorithm?
Limits With an Integral in the Numerator
Some limits have a definite integral with a variable upper limit on top. You don't need to evaluate the integral. The Fundamental Theorem of Calculus does the differentiating for you.
Check: here the integral actually equals $e^x-1-x$, and the graph of $\dfrac{e^x-1-x}{x^2}$ does approach $\frac12$ at $0$.
In a biotechnology laboratory, the growth of a bacteria population in a controlled environment follows a law whose cumulative accumulation rate is expressed by an…
In a biotechnology laboratory, the growth of a bacteria population in a controlled environment follows a law whose cumulative accumulation rate is expressed by an integral. It is desired to study the behavior of the population density at the initial instant by calculating the following limit:
$$\lim_{x \to 0} \frac{\int_{0}^{x} \ln(1 + t^2) \, dt}{x^3}$$Use L'Hôpital's Rule and the Fundamental Theorem of Calculus to determine the value of this limit.
Unknown Parameters
A classic exam problem: the function has letters like $a$ and $b$, and you are told the limit exists (or has a given value). Find the letters.
Drag the slider: only $a=-1$ makes the graph settle at a finite height near $x=0$.
Given the function dependent on the parameter $k \in \mathbb{R}$: $$f(x) = \frac{e^x - kx - 1}{x^2}$$ Calculate the value of $k$ so that the limit $\lim_{x \to 0} f(x)$…
Given the function dependent on the parameter $k \in \mathbb{R}$:
$$f(x) = \frac{e^x - kx - 1}{x^2}$$Calculate the value of $k$ so that the limit $\lim_{x \to 0} f(x)$ is a finite real number and determine this value. Explain why it is necessary for the expression in the numerator to be zero at $x=0$ to be able to apply the rule.
A team of sound engineers is analyzing the response of an audio filter. The gain of the filter as a function of the frequency $x$ is given by the expression: $$G(x) =…
A team of sound engineers is analyzing the response of an audio filter. The gain of the filter as a function of the frequency $x$ is given by the expression:
$$G(x) = \frac{a \cdot \cos(x) + b \cdot e^x - 2}{x^2}$$Knowing that the limit of the gain as the frequency approaches zero is equal to $1$, that is, $\lim_{x \to 0} G(x) = 1$, find the values of the parameters $a$ and $b$.
In the design of a new roller coaster, the curvature of an initial section is determined by the existence of the limit of a transition function. For the design to be…
In the design of a new roller coaster, the curvature of an initial section is determined by the existence of the limit of a transition function. For the design to be viable, the following limit is required to be a finite real number:
$$\lim_{x \to 0} \frac{a \cdot \cos(x) + b \cdot e^x - 3 + x}{x^2}$$Find the values of the parameters $a, b \in \mathbb{R}$ so that this limit exists and is finite. Once obtained, calculate the value of this limit.
The Forms $0\cdot\infty$ and $\infty-\infty$
These forms can't be differentiated directly. Rewrite them as a single quotient first.
Calculate the value of the following limit by first transforming the indeterminate form of type $\infty - \infty$ into a quotient to be able to apply L'Hôpital's Rule:…
Calculate the value of the following limit by first transforming the indeterminate form of type $\infty - \infty$ into a quotient to be able to apply L'Hôpital's Rule:
$$\lim_{x \to 1} \left( \frac{x}{x-1} - \frac{1}{\ln(x)} \right)$$Calculate the limit of the following function as $x$ approaches $\frac{\pi}{2}$. Note that you will need to perform a prior transformation to convert the indeterminate…
Calculate the limit of the following function as $x$ approaches $\frac{\pi}{2}$. Note that you will need to perform a prior transformation to convert the indeterminate form $0 \cdot \infty$ into a form suitable for L'Hôpital's Rule:
$$\lim_{x \to \frac{\pi}{2}} \left( \sin(x) - 1 \right) \cdot \tan(x)$$A telecommunications engineer analyzes the power loss of a signal in an optical fiber. The relative power $P$ with respect to the distance $d$ is modeled by a function…
A telecommunications engineer analyzes the power loss of a signal in an optical fiber. The relative power $P$ with respect to the distance $d$ is modeled by a function that requires solving the following indeterminate form when the signal approaches a critical node ($d \to 1$):
$$\lim_{d \to 1} \left( \frac{d}{\ln(d)} - \frac{1}{d-1} \right)$$Calculate the value of the limit, indicating the initial type of indeterminate form and the algebraic transformations carried out to apply L'Hôpital's Rule.
The Form $1^\infty$
A power $f(x)^{g(x)}$ can be tricky because the base and the exponent pull in opposite directions. The trick is to take the natural log, which brings the exponent down.
On a social network, the number of shares of a viral video during the first few seconds has been modeled by the function $S(t) = (1+2t)^{1/t}$, where $t$ is the time in…
On a social network, the number of shares of a viral video during the first few seconds has been modeled by the function $S(t) = (1+2t)^{1/t}$, where $t$ is the time in seconds.
What value does the number of shares approach at the initial instant? To do this, calculate:
$$\lim_{t \to 0} (1+2t)^{\frac{1}{t}}$$Determine the value of the following limit which presents an indeterminate form of the type $1^\infty$. To solve it, use the properties of logarithms and L'Hôpital's…
Determine the value of the following limit which presents an indeterminate form of the type $1^\infty$. To solve it, use the properties of logarithms and L'Hôpital's Rule:
$$\lim_{x \to 0} (\cos x)^{\frac{1}{x^2}}$$Calculate the value of the following limit, which presents an indeterminate form of the type $1^\infty$. Make sure to transform the expression appropriately before…
Calculate the value of the following limit, which presents an indeterminate form of the type $1^\infty$. Make sure to transform the expression appropriately before applying L'Hôpital's Rule:
$$\lim_{x \to 0} \left( \frac{\sin(x)}{x} \right)^{\frac{1}{x^2}}$$The Forms $0^0$ and $\infty^0$
The logarithm method works for all the power forms. Here is the full list of indeterminate forms and how to handle them.
| Form | How to handle it |
|---|---|
| $\frac00$, $\frac\infty\infty$ | Apply L'Hôpital's Rule directly |
| $0\cdot\infty$ | Rewrite as a quotient |
| $\infty-\infty$ | Combine into one fraction |
| $1^\infty$, $0^0$, $\infty^0$ | Take $\ln$, then rewrite as a quotient |
Calculate the value of the limit of the following function as $x$ approaches zero from the right ($x \to 0^+$). Identify the type of indeterminate form and use the…
Calculate the value of the limit of the following function as $x$ approaches zero from the right ($x \to 0^+$). Identify the type of indeterminate form and use the relationship between powers and exponentials ($u^v = e^{v \cdot \ln(u)}$):
$$\lim_{x \to 0^+} (\tan(x))^x$$Continuity and Differentiability at a Point
L'Hôpital's Rule is often the tool for the key limit in a continuity or differentiability question. Recall that $f$ is continuous at $a$ if $\lim_{x\to a}f(x)$ exists and equals $f(a)$. Differentiability at $a$ requires the limit $f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ to exist.
Determine, if they exist, the values of the parameters $a$ and $b$ so that the following function is continuous throughout its domain: $$f(x) = \begin{cases}…
Determine, if they exist, the values of the parameters $a$ and $b$ so that the following function is continuous throughout its domain:
$$f(x) = \begin{cases} \dfrac{\sin(ax^2)}{x^2} & \text{if } x < 0 \\ b + 2 & \text{if } x = 0 \\ \dfrac{e^x - e^{-x} - 2x}{x - \sin(x)} & \text{if } x > 0 \end{cases}$$To solve the right-hand limit, it will be necessary to apply L'Hôpital's Rule repeatedly. Justify each step of the process.
Consider the function $g(x) = x^2 \cdot e^{-1/x^2}$ for $x \neq 0$. 1. Study whether it is possible to assign a value to $g(0)$ so that the function is continuous at…
Consider the function $g(x) = x^2 \cdot e^{-1/x^2}$ for $x \neq 0$.
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Study whether it is possible to assign a value to $g(0)$ so that the function is continuous at $x = 0$.
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Using the definition of the derivative as a limit and applying L'Hôpital's Rule, prove whether the function is differentiable at $x = 0$.
Note: Take special care with the change of variable needed to solve the limit at infinity if necessary.
Removable Discontinuities and Asymptotes
A removable discontinuity at $x=a$ means the limit exists (a finite number) but $f(a)$ is undefined or has the wrong value. For a rational function, this happens when the same factor $x-a$ is in both the numerator and the denominator, so both are $0$ at $a$. Limits at infinity tell you about asymptotes.
Given the function $f(x) = \frac{x^2 + kx + 2}{x^2 - 1}$, it is known that it has a removable discontinuity at $x = 1$. 1. Calculate the value of the constant $k$. 2.…
Given the function $f(x) = \frac{x^2 + kx + 2}{x^2 - 1}$, it is known that it has a removable discontinuity at $x = 1$.
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Calculate the value of the constant $k$.
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For this value of $k$, calculate the value the function should take at $f(1)$ for it to be continuous at that point, by solving the limit:
- Is there any slant asymptote for this function? Justify your answer using limits.
Back to the start: for $\dfrac{e^x-1}{\sin x}$, L'Hôpital's Rule gives $\dfrac{e^x}{\cos x}\to 1$, so the hole in the Desmos graph sits at height $1$.