L'Hôpital's Rule II

Politicians are not born; they are excreted, Marcus Tullius Cicero.

Recall

L’Hôpital’s Rule is a mathematical technique used to evaluate limits of indeterminate forms using derivatives. It states that for functions f and g which are differentiable on an interval I except possible at a point “a” contained in I, if $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ or ± ∞, and g'(x) ≠ 0 ∀x ∈ I, except possible at a, then $\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)},$ provided that the right hand limit exists or equals ± ∞.

This is the second article in our two-part series about the L’Hôpital’s Rule and is a continuation of the first one, so if you haven’t taken a look at it yet, I recommend you read it first and then, come back.

Solved exercises

• $\lim_{x \to ∞} \frac{x^2-9x+4}{3x^2+7x+8}=$ [L’Hôpital’s Rule, ∞/∞] = $\lim_{x \to ∞} \frac{2x-9}{6x+7} = \lim_{x \to ∞} \frac{2}{6} = \frac{1}{3}.$

• $\lim_{x \to ∞} \frac{\sqrt{x}}{e^x} =$[L’Hôpital’s Rule, ∞/∞] = $\lim_{x \to ∞} \frac{\frac{1}{2}·\frac{1}{\sqrt{x}}}{e^x} = \lim_{x \to ∞} \frac{\frac{1}{2\sqrt{x}}}{e^x} = \lim_{x \to ∞} \frac{1}{2\sqrt{x}e^x} = 0.$

• $\lim_{x \to -∞} x^2e^x$ [∞·0 is an indeterminate form] = $\lim_{x \to -∞} \frac{x^2}{e^{-x}} =$ [L’Hôpital’s Rule, ∞/∞] $\lim_{x \to -∞} \frac{2x}{-e^{-x}}$ [L’Hôpital’s Rule, -∞/-∞] $\lim_{x \to -∞} \frac{2}{e^{-x}} = 0.$ In words, as x tends to negative infinite, x² approaches positive infinity, but at a decreasing rate compared to e^x.

• $\lim_{x \to 2} \frac{ln(x^3-8)}{ln(x^2-3x+2)}$=[L’Hôpital’s Rule, -∞/-∞] $\lim_{x \to 2} \frac{\frac{1}{x^3-8}·3x^2}{\frac{1}{x^2-3x+2}·(2x-3)} = \lim_{x \to 2} \frac{\frac{3x^2}{x^3-8}}{\frac{2x-3}{x^2-3x+2}} = \lim_{x \to 2} \frac{3x^2(x^2-3x+2)}{(x^3-8)(2x-3)}$ =[Recall a³-b³ = (a-b)(a²+ab+b²)] =$\lim_{x \to 2} \frac{3x^2(x-1)(x-2)}{(x-2)(x^2+2x+4)(2x-3)} = \lim_{x \to 2} \frac{3x^2(x-1)}{(x^2+2x+4)(2x-3)} = \frac{3·4·1}{(4+4+4)·1} = 1.$

• $\lim_{x \to 0+} x^x = \lim_{x \to 0+} e^{ln(x^x)} = \lim_{x \to 0+} e^{x·ln(x)}$. Notice that $\lim_{x \to 0+} xln(x) = \lim_{x \to 0+} \frac{ln(x)}{1/x}$ =[L’Hôpital’s Rule, -∞/-∞] $\lim_{x \to 0+} \frac{1/x}{-1/x^2} = \lim_{x \to 0+} \frac{-x^2}{x} = \lim_{x \to 0+} -x = 0 ⇒ \lim_{x \to 0+} x^x = \lim_{x \to 0+} e^{xln(x)} = e^{\lim_{x \to 0+} xln(x)} = e^0 = 1.$

• Let a > 0, $\lim_{x \to ∞} xe^{-ax} = \lim_{x \to ∞} \frac{x}{e^{ax}}$ =[L’Hôpital’s Rule, ∞/∞] $\lim_{x \to ∞} \frac{1}{ae^{ax}} = \frac{1}{∞} = 0.$ In words, x grows slower than e^ax as x → ∞.

• Let a > 1, $\lim_{x \to ∞} \frac{e^{px}}{x^a} = (\lim_{x \to ∞} \frac{e^{px/a}}{x})^a$ =[L’Hôpital’s Rule] $(\lim_{x \to ∞} \frac{p}{a}\frac{e^{px/a}}{1})^a = (\frac{∞}{1})^a = ∞.$ In words, the exponential e^px grows faster than any power of x as x → ∞.

• $\lim_{x \to 0} \frac{x-sin(x)}{x-tan(x)}$ =[L’Hôpital’s Rule, 0/0] $\lim_{x \to 0} \frac{1-cos(x)}{1-sec^2(x)}$ =[It is still an indeterminate form 0/0, let’s multiply numerator and denominator by cos²(x)] $\lim_{x \to 0} \frac{(1-cos(x))cos^2(x)}{(1-sec^2(x))cos^2(x)} = \lim_{x \to 0} \frac{(1-cos(x))cos^2(x)}{cos^2(x)-1} = \lim_{x \to 0} \frac{(1-cos(x))cos^2(x)}{(cos(x)-1)(cos(x)+1)} = \lim_{x \to 0} \frac{-cos^2(x)}{cos(x)+1} = \frac{-1}{2}.$

• $\lim_{x \to ∞} \frac{ln(x)}{x^{1/3}}$ =[L’Hôpital’s Rule, ∞/∞] $\lim_{x \to ∞} \frac{1/x}{1/3·x^{-2/3}} =\lim_{x \to ∞} 3x^{2/3-1} =\lim_{x \to ∞} 3x^{-1/3} = 0.$ In words, ln(x) grows slower than $x^{1/3}$ or any arbitrary positive power of x.

• $\lim_{x \to ∞} x^{\frac{1}{x}} = \lim_{x \to ∞} e^{ln(x^{\frac{1}{x}})} = \lim_{x \to ∞} e^{ln(x)\frac{1}{x}} = e^{\lim_{x \to ∞} \frac{ln(x)}{x}}$ =[L’Hôpital’s Rule, ∞/∞] $e^{\lim_{x \to ∞} \frac{\frac{1}{x}}{1}} = e^{\lim_{x \to ∞} \frac{1}{x}} = e^0 = 1.$

• $\lim_{x \to ∞} (1-\frac{1}{x})^x =$ [1^∞ is an indeterminate form, it is not 1 💣, e.g., $\lim_{n \to ∞} (1+\frac{1}{n})^n = e ≈ 2.71828$] = $e^{\lim_{x \to ∞} ln((1-\frac{1}{x})^x)} = e^{\lim_{x \to ∞} x·ln(1-\frac{1}{x})} = e^{\lim_{x \to ∞} \frac{ln(1-\frac{1}{x})}{\frac{1}{x}}}$ =[L’Hôpital’s Rule, 0/0] $e^{\lim_{x \to ∞} \frac{\frac{1}{1-\frac{1}{x}}·\frac{1}{x^2}}{\frac{-1}{x^2}}} = e^{\lim_{x \to ∞} \frac{-1}{1-\frac{1}{x}}} = e^{\lim_{x \to ∞} \frac{-1}{\frac{x-1}{x}}} = e^{\lim_{x \to ∞} \frac{-x}{x-1}} = e^{\lim_{x \to ∞} \frac{-1}{1}} = e^{-1}= \frac{1}{e}.$

• $\lim_{x \to 0} (cos(x))^{\frac{1}{x^2}}$ =[1^∞ is an indeterminate form] $\lim_{x \to 0} e^{ln(cos(x))^{\frac{1}{x^2}}} = \lim_{x \to 0} e^{\frac{ln(cos(x))}{x^2}} = e^{\lim_{x \to 0} \frac{ln(cos(x))}{x^2}}$ [L’Hôpital’s Rule, 0/0] $e^{\lim_{x \to 0} \frac{\frac{1}{cos(x)}·-sin(x)}{2x}} = e^{\lim_{x \to 0} \frac{\frac{-sin(x)}{cos(x)}}{2x}}$ =[L’Hôpital’s Rule, 0/0] $e^{\lim_{x \to 0} \frac{\frac{-cos^2(x)-(-sin(x)(-sin(x)))}{cos^2(x)}}{2}} = e^{\lim_{x \to 0} \frac{\frac{-cos^2(x)-sin^2(x)}{cos^2(x)}}{2}} = e^{\lim_{x \to 0} \frac{\frac{-1}{cos^2(x)}}{2}} = e^{\frac{-1}{2}} = \frac{1}{\sqrt{e}}.$

Asymptotic Complexity

When analyzing the running time or space usage of programs, we usually try to estimate the time or space as function of the input size. For example, when analyzing the worst case running time of an algorithm that sorts a list of numbers, we will be concerned with how long it takes a computer to run the lines of code of the algorithm as a function of the length of the input list.

In mathematics and computer science, the concept of rates of growth (asymptotic growth) often refers to the asymptotic behavior of functions as their input values become (very) large and we express it as f(x) «_x→∞ g(x). It means $\lim_{x \to ∞}\frac{f(x)}{g(x)} = 0$, e.g., 1 « log₂(x) « x « xlog₂(x) « x²; ln(x) « x^p « e^x « $e^{x^2}$ as x → ∞, p > 0.

Analogously, 1/ln(x) » 1/x^p » e^-x » $e^{-x^2}$ as x → ∞, p > 0.

Big-O notation represents the upper bound of the running time of an algorithm, i.e., it gives the worst-case complexity of an algorithm. Constant factors are very small in the face of arbitrarily large n, so we ignored them, 7n is represented by O(n). In computation, big O notation is used to classify algorithms based on how their execution time or space requirements grow as the size of the input increases.

Asymptotic growth is categorized into several classes (complexity classes): constant (1) – not literally 1 but some constant, logarithmic (log₂(n)), linear (n), n·log(n), quadratic (n²), cubic (n³), and exponential (mⁿ where m is some constant): O(1) < O(log(n)) < O(n) < O(n·log(n)) < O(n²) < O(2ⁿ).

Bibliography

1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn, and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.