An expert is a person who has made all the mistakes that can be made in a very narrow field, Neils Bohr.

$\lim_{x \to 1} \frac{x^3-1}{x^2-1} =$[Indeterminate form, 0/0] = $\lim_{x \to 1} \frac{\frac{x^3-1}{x-1}}{\frac{x^2-1}{x-1}}$ [Notice that the numerator is just $\frac{x^3-1}{x-1} = \frac{f(x)-f(1)}{(x-1)}$ →_{x→1} f’(1) where f(x) = x^{3}-1, f(1) = 0] $\frac{3x^2}{2x} = \frac{3}{2}.$

More generally, let’s assume that f(a) = g(a) = 0, $\lim_{ x \to a}\frac{f(x)}{g(x)} = \lim_{ x \to a}\frac{f(x)/(x-a)}{g(x)/(x-a)}$ =[By assumption, f(a) = g(a) = 0] $\frac{\lim_{ x \to a}\frac{f(x)-f(a)}{(x-a)}}{\lim_{ x \to a}\frac{g(x)-g(a)}{(x-a)}} = \frac{f’(a)}{g’(a)},$ and that works provided that g’(a) ≠ 0.

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 ± ∞.

$\lim_{x \to 0}\frac{sin(3x)}{sin(2x)}$ =[L’Hôpital’s Rule] $\lim_{x \to 0}\frac{3cos(3x)}{2cos(2x)} =\frac{3cos(0)}{2cos(0)}=\frac{3}{2}.$

When dealing with limits and facing difficulties in directly evaluating them, approximation methods can be deployed to estimate the limit, e.g., sin(u) ≈ u ↭ u ≈ 0 ⇒ $\frac{3cos(3x)}{2cos(2x)}≈\frac{3x}{2x}=\frac{3}{2}.$ Analogously, $\frac{cos(x)-1}{x^2} ≈_{x≈0}\frac{(1-\frac{x^2}{2})-1}{x^2}=\frac{-x^2/2}{x^2}=\frac{-1}{2}.$

$\lim_{x \to 0}\frac{cos(x)-1}{x^2}$ =[L’Hôpital’s Rule] $\lim_{x \to 0}\frac{-sin(x)}{2x}$ = [There may be instances where we would need to apply L’Hôpital’s Rule multiple times] $\lim_{x \to 0}\frac{-cos(x)}{2} = \frac{-1}{2}.$

$\lim_{x \to 0^+} xln(x)$ =[It seems that L’Hôpital’s Rule does not apply, 0·(-∞)] $\lim_{x \to 0^+} \frac{ln(x)}{1/x}$ = [L’Hôpital’s Rule] $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} -x = 0$

$\lim_{x \to ∞} xe^{-px} = \lim_{x \to ∞} \frac{x}{e^{px}}$ =[L’Hôpital’s Rule, ∞/∞] $\lim_{x \to ∞} \frac{1}{pe^{px}} = \frac{1}{∞} = 0.$ In words, x grows slower than e^{px} 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 → ∞.

$\lim_{x \to ∞} \frac{ln(x)}{x^{1/3}}$ =[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.

$\lim_{x \to 0+} x^x$ =[Recall that a^{b}=e^{bln(a)}] $\lim_{x \to 0+} e^{xln(x)}$. Notice that $\lim_{x \to 0+} xln(x) = \lim_{x \to 0+} \frac{ln(x)}{1/x} = \lim_{x \to 0+} \frac{1/x}{-1/x^2} = \lim_{x \to 0+} -x = 0 ⇒ \lim_{x \to 0+} e^{xln(x)} = e^0 = 1.$

$\lim_{x \to 0+} \frac{sin(x)}{x^2} = \lim_{x \to 0+} \frac{cos(x)}{2x} = \lim_{x \to 0+} \frac{-sin(x)}{2} = 0$. However, using the approximation method sin(x) ≈ x, hence sin(x)/x^{2} ≈ 1/x → ∞. There is something very fishy🐟🐬, what is it?

This is false because cos(x)/2x indeterminate is of the form “1/0”, and therefore, we are applying L’Hôpital’s Rule wrong, **$\lim_{x \to 0+} \frac{cos(x)}{2x} ≠ \lim_{x \to 0+} \frac{-sin(x)}{2}$**.

In mathematics and computer science, the concept of rates of 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., 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.

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License and is based on MIT OpenCourseWare [18.01 Single Variable Calculus, Fall 2007].

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