L'Hôpital's Rule

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

Recall

Very loosing speaking, a limit is the value to which a function grows close as the input get closer and closer to some other given value.

A limit involves what is going on around a given point, say “a”. It is the value that a function approaches as its input approaches that point “a” but does not care what the function is at this particular point, f(a).

Limits are essential to calculus and mathematical analysis and the understanding of how functions behave. The concept of a limit can be written or expressed as: $\lim_{x \to a} f(x) = L.$ This is read as “the limit of f as x approaches a equals L”. This means that the value of the function f can be made arbitrarily close to L (and I mean as close as you want, e.g., L ± 0.1, L ± 0.01, L ± 0.001, you get the idea 😄), by choosing values of x sufficiently close to a.

There are several basic methods used to solve limits: direct substitution (the first thing you should always try when calculating limits, is just entering the x value into the function, e.g., $\lim_{x \to 3}(2x^2-4x+1) = 2·3^2-4·3+1 = 18-12+1 = 7$), factorization (this is basically a technique to finding limits that works by cancelling out common factors, e.g., $\lim_{x \to 1} \frac{x^2-2x+1}{x-1} = \lim_{x \to 1} \frac{(x-1)^2}{(x-1)} = \lim_{x \to 1} (x-1) = 0$), and L’Hopital’s rule.

The indeterminate forms are forms whose value cannot be determined when we evaluate limits by substituting the x-value into the function: 0/0, 0·∞, ∞/∞, ∞ − ∞, ∞⁰, 0⁰, and 1^∞. L’Hopital’s rule is a mathematical technique used to evaluate limits of indeterminate forms, such as 0/0 or infinity/infinity.

Motivation

$\lim_{x \to 1} \frac{x^3-1}{x^2-1} =$[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) = x³-1, f(1) = 0. Similarly, we can argue about the denominator, g(x) = x²-1] $\lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{3x^2}{2x} = \frac{3}{2}.$

More generally, let f and g be differentiable functions, f’ and g’ continuous, 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 ± ∞.

Solved exercises

• $\lim_{x \to 0}\frac{e^x-1}{x^2+x}$ =[L’Hôpital’s Rule, indeterminate form 0/0] = $\lim_{x \to 0}\frac{e^x}{2x+1} = 1.$

• $\lim_{x \to ∞}\frac{3x^2-2x+1}{4x^3-6x+1}$ [L’Hôpital’s Rule, indeterminate form ∞/∞] = $\lim_{x \to ∞}\frac{6x-2}{12x^2-6}$ [L’Hôpital’s Rule, indeterminate form ∞/∞] = $\lim_{x \to ∞}\frac{6}{24x} = 0.$

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

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 ≈ 0} u ⇒ $\frac{sin(3x)}{sin(2x)}≈\frac{3x}{2x}=\frac{3}{2}.$ 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, indeterminate form 0/0] $\lim_{x \to 0}\frac{-sin(x)}{2x}$ = [There may be instances where we would need to apply L’Hôpital’s Rule multiple times] $\lim_{x \to 0}\frac{-cos(x)}{2} = \frac{-1}{2}.$

• $\lim_{x \to 0^+} (\frac{1}{x}-\frac{1}{1-cos(x)}) =$[We are evaluating a limits of an indeterminate form. It seems that L’Hôpital’s Rule does not apply.] $ \lim_{x \to 0^+} (\frac{1-cos(x)-x}{x-xcos(x)}) =$ [L’Hospital Rule, 0/0] $\lim_{x \to 0^+} \frac{sin(x)-1}{1-(cos(x)+x(-sin(x)))} = \lim_{x \to 0^+} \frac{sin(x)-1}{1-cos(x)+xsin(x)} = \frac{-1}{1-1+0}=-∞.$

• $\lim_{x \to 0^+} csc(x)·ln(1-sin(x)) =$[We are evaluating a limits of an indeterminate form, ∞·0. It seems that L’Hôpital’s Rule does not apply.] $\lim_{x \to 0^+} \frac{ln(1-sin(x))}{sin(x)}$ = [L’Hôpital’s Rule] $\lim_{x \to 0^+} \frac{\frac{1}{1-sin(x)}·-cos(x)}{cos(x)} = \lim_{x \to 0^+} \frac{-1}{1-sin(x)} = \frac{-1}{1} = -1.$

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

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

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

• $\lim_{x \to 0} \frac{x+sin(x)}{x+cos(x)} = \frac{0}{1} = 0$. We need to determine whether or not l’Hospital’s Rule applies (0/0, ∞/∞ indeterminates). If we were to use L'Hospital Rule here, we would actually get an incorrect answer.

• $\lim_{x \to 0+} \frac{sin(x)}{x^2} = \lim_{x \to 0+} \frac{cos(x)}{2x} = \lim_{x \to 0+} \frac{-sin(x)}{2} = 0$. However, using the approximation method sin(x) ≈ x, hence sin(x)/x² ≈ 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, $\lim_{x \to 0+} \frac{cos(x)}{2x} ≠ \lim_{x \to 0+} \frac{-sin(x)}{2}$.

Bibliography

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