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An expert is a person who has made all the mistakes that can be made in a very narrow field, Neils Bohr.
Very loosely speaking, a limit is the value to which a function grows close as the input gets 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 it does not care at all what the function actually is at this particular point, $f(a)$. The function might not even exist at $a$!
Limits are the very foundation of calculus and mathematical analysis. Without limits, we can’t define instantaneous rates of change (derivatives) or the exact accumulation of quantities (integrals).
The concept of a limit is mathematically expressed as: $\lim_{x \to a} f(x) = L$. This is read as “the limit of $f(x)$ as $x$ approaches $a$ equals $L$.”
The value of the function $f(x)$ can be made arbitrarily close to $L$ (and I mean as close as you want, e.g., $L \pm 0.1$, $L \pm 0.01$, $L \pm 0.001$ —I hope you get the idea 😄) simply by picking values of $x$ that are sufficiently close to $a$ (from either side, but never exactly $a$).
When you are asked to evaluate a limit, there is a natural hierarchy of strategies you should attempt.
Direct Substitution: The very first thing you should always try when calculating a limit is just plugging the $x$-value directly into the function. If the function is continuous at that point, you’re done!
Example: $\lim_{x \to 3} (2x^2 - 4x + 1) = 2(3)^2 - 4(3) + 1 = 18 - 12 + 1 = 7$.
Factorization: If substitution gives you a division by zero, check if you can cancel out the “problem” by factoring the expression. This technique reveals the hidden limit by removing the mathematical “hole” in the graph.
Example: $\lim_{x \to 1} \frac{x^2 - 2x + 1}{x - 1} = \lim_{x \to 1} \frac{(x - 1)^2}{x - 1} = \lim_{x \to 1} (x - 1) = 0$.
L’Hôpital’s Rule: When substitution yields nonsense (an indeterminate form) and algebra fails you, calculus rides to the rescue.
Indeterminate forms are expressions whose mathematical limits cannot be determined simply by substituting the $x$-value into the function. The classic culprits are:
L’Hôpital’s Rule is a brilliant mathematical technique used to evaluate limits of those first two primary indeterminate forms ($\frac{0}{0}$ and $\frac{\infty}{\infty}$). For the others, we usually have to do some algebraic manipulation to force them into a fraction first!
Let’s look at a concrete example: $\lim_{x \to 1} \frac{x^3 - 1}{x^2 - 1}$.
Plugging in $x = 1$ yields $\frac{0}{0}$, an indeterminate form.
But watch what happens if we cleverly manipulate the fraction by dividing the top and bottom by $(x-1)$: $\lim_{x \to 1} \frac{\frac{x^3 - 1}{x - 1}}{\frac{x^2 - 1}{x - 1}}$
Look closely at the numerator. It is literally the definition of the derivative of $f(x) = x^3$ at $x=1$: $\lim_{x \to 1} \frac{f(x) - f(1)}{x - 1} = f'(1)$
The same is true for the denominator with $g(x) = x^2$.
So, evaluating the limit of the ratio of the functions is the same as evaluating the ratio of their derivatives!
$\lim_{x \to 1} \frac{3x^2}{2x} = \frac{3}{2}$
More generally, let f, g be differentiable at a, with f(a) = g(a) = 0, and assume $g'(a) \neq 0$. Then, for $x\neq a$, $\frac{f(x)}{g(x)} = \frac{\dfrac{f(x)-f(a)}{x-a}}{\dfrac{g(x)-g(a)}{x-a}}.$
Taking the limit as $x \to a$, and using the fact that f’ and g’ are continuous at a (or at least that the limits of the difference quotients exist and equal f’(a) and g’(a) respectively), we get $\lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}}{\lim_{x\to a}\dfrac{g(x)-g(a)}{x-a}} = \frac{f'(a)}{g'(a)}$, provided the denominator limit (g’(a)) is nonzero.
Let $f$ and $g$ be differentiable functions on an interval $I$, except possibly at a point $a$ contained in $I$. If $\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0$ (or $\pm \infty$), and $g'(x) \neq 0$ for all $x \in I$ (except possibly at $a$), then: $\boxed{\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}}$, and that works provided that the limit on the right-hand side actually exists or equals $\pm \infty$.
Let’s look at how L’Hôpital’s rule tackles different flavors of indeterminate forms.
$\lim_{x \to 0} \frac{e^x - 1}{x^2 + x}$
Check: Plug in 0 → $\frac{1-1}{0} = \frac{0}{0}$.
L’Hôpital: $\lim_{x \to 0} \frac{\frac{d}{dx}(e^x - 1)}{\frac{d}{dx}(x^2 + x)} = \lim_{x \to 0} \frac{e^x}{2x + 1} = \frac{1}{1} = 1.$
$\lim_{x \to \infty} \frac{3x^2 - 2x + 1}{4x^3 - 6x + 1}$
Check: As $x \to \infty$, we get $\frac{\infty}{\infty}$.
L’Hôpital (1st time): $\lim_{x \to \infty} \frac{6x - 2}{12x^2 - 6}$ (Still $\frac{\infty}{\infty}$)
L’Hôpital (2nd time): $\lim_{x \to \infty} \frac{6}{24x} = 0.$
When you encounter forms like $0 \cdot \infty$ or $\infty - \infty$, you cannot apply L’Hôpital’s rule immediately. You must rewrite them as fractions first.
($0 \cdot -\infty$): $\lim_{x \to 0^+} x \ln(x)$
Check: $0 \cdot (-\infty)$.
Transformation: Rewrite $x$ as $\frac{1}{1/x}$. Now we have $\lim_{x \to 0^+} \frac{\ln(x)}{1/x}$, which is $\frac{-\infty}{\infty}$.
L’Hôpital: $\lim_{x \to 0^+} \frac{1/x}{-1/x^2} = \lim_{x \to 0^+} (-x) = 0.$
($\infty - \infty$): $\lim_{x \to 1^+} \left( \frac{x}{x - 1} - \frac{1}{\ln(x)} \right)$
Check: $\infty - \infty$. Find a common denominator to force a fraction.
Transformation: $\lim_{x \to 1^+} \frac{x\ln(x) - (x - 1)}{(x - 1)\ln(x)}$. This is now $\frac{0}{0}$.
L’Hôpital (1st time): $\lim_{x \to 1^+} \frac{\ln(x) + x(\frac{1}{x}) - 1}{\ln(x) + (x - 1)(\frac{1}{x})} = \lim_{x \to 1^+} \frac{\ln(x)}{\ln(x) + 1 - \frac{1}{x}}$. (Still $\frac{0}{0}$).
L’Hôpital (2nd time): $\lim_{x \to 1^+} \frac{1/x}{1/x + 1/x^2}$. Plug in $x=1 \to \frac{1}{1 + 1} = \frac{1}{2}.$
When dealing with trigonometric limits near zero, Taylor series approximations can often bypass L’Hôpital entirely! For $x \approx 0$:
So, instead of L’Hôpital, you could evaluate $\lim_{x \to 0} \frac{\sin(3x)}{\sin(2x)}$ simply as $\frac{3x}{2x} = \frac{3}{2}$. Similarly: $\lim_{x \to 0} \frac{\cos(x) - 1}{x^2} \approx \frac{(1 - x^2/2) - 1}{x^2} = \frac{-x^2/2}{x^2} = -\frac{1}{2}$.
Obviously, $\lim_{x \to 0}\frac{cos(x)-1}{x^2}$ =[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] $\lim_{x \to 0}\frac{-cos(x)}{2} = \frac{-1}{2}.$. Similarly, $\lim_{x \to 0}\frac{sin(3x)}{sin(2x)}$ =[L’Hôpital’s Rule, indeterminate form 0/0] $\lim_{x \to 0}\frac{3cos(3x)}{2cos(2x)} =\frac{3cos(0)}{2cos(0)}=\frac{3}{2}.$
L’Hôpital’s Rule is powerful, but it is not a magic wand for every fraction. You must verify the indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) first!
The False Application:
If you blindly apply L’Hôpital’s Rule without checking: $\lim_{x \to 0} \frac{1 + \cos(x)}{1 - \sin(x)} = \frac{1 + 1}{1 - 0} = 2$.
This is completely wrong. Why? Because if we just use direct substitution initially: $\frac{0 + \sin(0)}{0 + \cos(0)} = \frac{0}{1} = 0$.
Because the limit was not indeterminate to begin with, L’Hôpital’s Rule yields a false result.
Consider $\lim_{x \to 0^+} \frac{\sin(x)}{x^2}$.
If we try L’Hôpital: $\lim_{x \to 0^+} \frac{\cos(x)}{2x}$.
If we apply L’Hôpital a second time (without checking!): $\lim_{x \to 0^+} \frac{-\sin(x)}{2} = 0$.
There is something very fishy 🐟🐬 here. If we use the approximation $\sin(x) \approx x$, we get $\frac{x}{x^2} = \frac{1}{x}$, which clearly shoots to $\infty$ as $x \to 0^+$.
What went wrong? The step where we got $\frac{\cos(x)}{2x}$ evaluates to $\frac{1}{0}$. This is not an indeterminate form —it’s a massive clue that the limit approaches infinity! Because $\frac{1}{0}$ is not $\frac{0}{0}$ or $\frac{\infty}{\infty}$, applying L’Hôpital that second time was illegal.
$\lim_{x \to 0^+} (\frac{1}{x}-\frac{1}{1-\cos(x)}) =$[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 - x\cos(x)}) =$ [L’Hospital Rule, 0/0] $\lim_{x \to 0^+} \frac{\sin(x)-1}{1-(\cos(x)+x(-\sin(x)))} = \lim_{x \to 0^+} \frac{sin(x)-1}{1-\cos(x) + x\sin(x)} = \frac{-1}{1-1+0}=-\infty.$
$\lim_{x \to 0^+} \csc(x) \cdot \ln(1-\sin(x)) =$[We are evaluating a limits of an indeterminate form, ∞·0. It seems that L’Hôpital’s Rule does not apply. Rewrite the product] $\lim_{x \to 0^+} \frac{\ln(1-\sin(x))}{\sin(x)}$ = [L’Hôpital’s Rule] $\lim_{x \to 0^+} \frac{\frac{1}{1-\sin(x)}\cdot-\cos(x)}{\cos(x)} = \lim_{x \to 0^+} \frac{-1}{1-\sin(x)} = \frac{-1}{1} = -1.$
$\lim_{x \to 0^+} (\frac{1}{sin(x)}-\frac{1}{x})$ [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] $\lim_{x \to 0^+} \frac{-(-sin(x))}{cos(x)+cos(x)+x(-sin(x))} = \lim_{x \to 0^+} \frac{-sin(x)}{2cos(x)-xsin(x)} = \frac{0}{2} = 0.$
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