Target Audience: Mathematics students, engineers, and self-learners.
Prerequisites: Basic algebra and trigonometry.
Before we memorize formulas, we must understand the question a limit asks. The expression:
asks: "As x gets arbitrarily close to a (without actually being a), what value does f(x) get close to?"
Why Standard Limits Matter: In arithmetic, we learn multiplication tables \(7 \times 8 = 56\) so we don't have to add 7 eight times. In Calculus, "Standard Limits" are the building blocks. They allow us to solve complex problems by breaking them down into recognizable patterns.
For "well-behaved" functions (continuous functions), the limit is simply the function value.
The Rule: For any polynomial P(x) and any real number a:
$$ \lim_{x \to a} P(x) = P(a) $$Specific Case:
$$ \lim_{x \to a} x^n = a^n $$Example 1: Evaluate \(\lim_{x \to 2} (3x^2 + 5x - 1)\).
The function is continuous, so the limit is simply the y-value at x=2.
Consider \(f(x) = \frac{x^2 - 1}{x - 1}\). If we plug in x = 1, we get \(\frac{0}{0}\), which is undefined. However, look at the graph below. The line looks exactly like y = x + 1, except for a single "hole" at x = 1.
Notice that while f(1) does not exist, the path approaches y=2.
Limits describe the path, not the destination. As x approaches 1 from the left (0.9, 0.99) and right (1.1, 1.01), the value approaches 2.
Note for Beginners: Understanding discontinuity types helps explain why limits exist or fail to exist. This foundational knowledge is essential for the standard limits we'll study in later sections. Feel free to revisit this section as your understanding deepens.
A removable discontinuity (also called a "hole") occurs when:
Key characteristic: You can "remove" or "fix" this discontinuity by redefining the function at that single point.
Our example: \(f(x) = \frac{(x-1)(x+1)}{x-1}\) has a removable discontinuity at x=1 because \(\lim_{x \to 1} f(x) = 2\) exists, even though f(1) is undefined. We could "fix" it by defining f(1) = 2.
These cannot be fixed by simply redefining a single point. There are several types:
The left and right limits exist but are different.
At x=0: \(\lim_{x \to 0^-} = 1\), but \(\lim_{x \to 0^+} = 2\). The function "jumps."
The function approaches \(\pm\infty\) (vertical asymptote).
At x=2: the function goes to \(\pm\infty\). The limit does not exist.
The function oscillates wildly with no limit.
At x=0: oscillates infinitely fast. No limit exists.
Removable
f(x) = x + 1 with hole
Jump
Piecewise function
Infinite
f(x) = 1/x
Essential
\(f(x) = \sin(1/x)\)
Every standard limit in this section shares a critical property: the function is undefined at the limit point. When we try direct substitution, we get meaningless expressions like \(\frac{0}{0}\).
Example: \(f(x) = \frac{\sin x}{x}\)
This is the power of limits: They tell us what value a function approaches, even when the function itself doesn't exist at that point.
An indeterminate form is an expression that doesn't have a determined value without further analysis. The most common in standard limits is \(\frac{0}{0}\), but there are seven classical forms:
Most common
See Euler's limit
Each standard limit below involves a function that is undefined at \(x = 0\) but has a well-defined limit:
| Limit Expression | At x = 0 | Limit Value |
|---|---|---|
| \(\frac{\sin x}{x}\) | \(\frac{0}{0}\) undefined | 1 |
| \(\frac{\tan x}{x}\) | \(\frac{0}{0}\) undefined | 1 |
| \(\frac{1 - \cos x}{x}\) | \(\frac{0}{0}\) undefined | 0 |
| \(\frac{1 - \cos x}{x^2}\) | \(\frac{0}{0}\) undefined | \(\frac{1}{2}\) |
| \(\frac{e^x - 1}{x}\) | \(\frac{0}{0}\) undefined | 1 |
| \(\frac{\ln(1+x)}{x}\) | \(\frac{0}{0}\) undefined | 1 |
Connection to Level 1: Removable Discontinuities
Remember the "hole" in \(f(x) = \frac{x^2-1}{x-1}\) at x=1? Each standard limit function has a similar "hole" at x=0. The limit tells us exactly where that hole would be filled — the value the function wants to reach but cannot.
(Note: x must be in radians)
Geometric Intuition: Imagine a very small arc on a unit circle. As the angle x becomes tiny, the length of the arc (x), the chord (\(\approx 2\sin(x/2)\)), and the vertical height (\(\sin x\)) become nearly indistinguishable. The ratio of the sine height to the arc length approaches 1.
Move the slider toward 0° to see sin(x)/x approach 1
Geometric Proof (Complete): For \(0 \lt x \lt \pi/2\), we compare areas on the unit circle:
Area Comparison:
Step 1: Establish the inequality chain
Since Inner Triangle \(\subseteq\) Sector \(\subseteq\) Outer Triangle:
$$ \frac{1}{2}\sin x \le \frac{x}{2} \le \frac{1}{2}\tan x $$Step 2: Multiply all parts by \(\frac{2}{\sin x}\) (positive for \(0 \lt x \lt \pi/2\))
$$ 1 \le \frac{x}{\sin x} \le \frac{1}{\cos x} $$Step 3: Take reciprocals (reverses inequalities)
$$ \cos x \le \frac{\sin x}{x} \le 1 $$Step 4: Apply limits as \(x \to 0^+\)
$$ \lim_{x \to 0^+} \cos x = 1 \quad \text{and} \quad \lim_{x \to 0^+} 1 = 1 $$Step 5: By the Squeeze Theorem
Since \(\cos x \le \frac{\sin x}{x} \le 1\) and both bounds approach 1:
$$ \lim_{x \to 0^+} \frac{\sin x}{x} = 1 $$Extension to Negative x: The function \(f(x) = \frac{\sin x}{x}\) is an even function because both \(\sin x\) and x are odd functions, making their ratio even: \(\frac{\sin(-x)}{-x} = \frac{-\sin x}{-x} = \frac{\sin x}{x}\). Therefore:
$$ \lim_{x \to 0^-} \frac{\sin x}{x} = \lim_{x \to 0^+} \frac{\sin x}{x} = 1 $$Thus, the two-sided limit exists: \(\displaystyle\lim_{x \to 0} \frac{\sin x}{x} = 1\)
Notice: The function approaches 1 as \(x \to 0\), and the oscillations dampen as \(|x|\) increases. The horizontal dashed line shows y = 1 (the limit value).
Derivation: Multiply by the conjugate \((1+\cos x)\):
$$ \lim_{x \to 0} \frac{1 - \cos x}{x} \cdot \frac{1 + \cos x}{1 + \cos x} = \lim_{x \to 0} \frac{\sin^2 x}{x(1+\cos x)} $$ $$ = \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0} \frac{\sin x}{1+\cos x} \right) = 1 \cdot \frac{0}{2} = 0 $$Derivation: We decompose the tangent function using \(\tan x = \frac{\sin x}{\cos x}\):
$$ \frac{\tan x}{x} = \frac{\sin x}{x \cos x} = \frac{\sin x}{x} \cdot \frac{1}{\cos x} $$Taking the limit as \(x \to 0\) and using our fundamental limit:
$$ \lim_{x \to 0} \frac{\tan x}{x} = \left( \lim_{x \to 0} \frac{\sin x}{x} \right) \cdot \left( \lim_{x \to 0} \frac{1}{\cos x} \right) = 1 \cdot \frac{1}{1} = 1 $$Derivation Method 1 (Half-Angle Identity):
Recall the half-angle identity: \(1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)\)
$$ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \lim_{x \to 0} \frac{2\sin^2(x/2)}{x^2} = \lim_{x \to 0} \frac{2\sin^2(x/2)}{4(x/2)^2} $$ $$ = \frac{1}{2} \cdot \lim_{x \to 0} \left( \frac{\sin(x/2)}{x/2} \right)^2 = \frac{1}{2} \cdot (1)^2 = \frac{1}{2} $$Derivation Method 2 (Conjugate Multiplication):
$$ \frac{1 - \cos x}{x^2} = \frac{(1 - \cos x)(1 + \cos x)}{x^2(1 + \cos x)} = \frac{\sin^2 x}{x^2(1 + \cos x)} $$ $$ = \left(\frac{\sin x}{x}\right)^2 \cdot \frac{1}{1 + \cos x} \xrightarrow{x \to 0} (1)^2 \cdot \frac{1}{1+1} = \frac{1}{2} $$Application Note: This limit appears frequently in Taylor series expansions and when evaluating derivatives of trigonometric functions. It's essential for understanding the behavior of \(\cos x \approx 1 - \frac{x^2}{2}\) near x = 0.
The number e is defined by a limit representing continuous growth:
$$ \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e $$A natural first reaction: as \(n \to \infty\), the base \(1 + \frac{1}{n} \to 1\), and 1 raised to any power is 1. So why isn't the limit just 1?
The tempting reasoning is:
$$ 1 + \frac{1}{n} \longrightarrow 1 \quad\text{as } n\to\infty, \qquad\text{so}\qquad \left(1+\tfrac{1}{n}\right)^n \longrightarrow 1^\infty = 1. $$The flaw is in the last step. \(1^\infty\) is an indeterminate form — it is not equal to 1. You cannot evaluate the limit by taking the limit of the base first and then applying the exponent. Both the base and the exponent are changing simultaneously, and they are fighting each other:
The whole question is: who wins? The answer is neither completely — they balance into a finite number, \(e \approx 2.71828\).
For any finite \(n\), the base is strictly greater than 1, so you are raising a number \(\gt 1\) to an ever-larger power. The extra \(\frac{1}{n}\) shrinks, but you apply it \(n\) times. Watch the tug-of-war numerically:
| \(n\) | \(1+\frac{1}{n}\) | \(\left(1+\frac{1}{n}\right)^n\) |
|---|---|---|
| 1 | 2.0 | 2.0 |
| 10 | 1.1 | 2.5937… |
| 100 | 1.01 | 2.7048… |
| 1,000 | 1.001 | 2.7169… |
| 1,000,000 | 1.000001 | 2.7182804… |
The base marches toward 1, but the power climbs toward \(e\) — not 1.
Your instinct works when the exponent is fixed:
$$ \left(1+\tfrac{1}{n}\right)^{5} \longrightarrow 1^5 = 1. $$It also fails the other direction if the exponent grows too slowly or the base shrinks too fast. The limit is finite and equal to \(e\) precisely because the "amount above 1" (\(\frac{1}{n}\)) and the exponent (\(n\)) are matched in size — their product \(n \cdot \frac{1}{n} = 1\) stays constant.
Take logarithms — this converts the indeterminate \(1^\infty\) into a workable \(\frac{0}{0}\):
$$ \ln\!\left(1+\tfrac1n\right)^n = n\ln\!\left(1+\tfrac1n\right) = \frac{\ln\!\left(1+\frac1n\right)}{\frac1n}. $$Let \(h = \frac{1}{n} \to 0\). The expression becomes
$$ \frac{\ln(1+h) - \ln(1)}{h}, $$which is exactly the definition of the derivative of \(\ln x\) at \(x = 1\). Since \(\frac{d}{dx}\ln x = \frac{1}{x}\), that derivative is \(\frac{1}{1} = 1\). So
$$ \ln\!\left(1+\tfrac1n\right)^n \longrightarrow 1 \quad\Longrightarrow\quad \left(1+\tfrac1n\right)^n \longrightarrow e^1 = e. $$The logarithm of your expression approaches 1 — that's where the "1" you were sensing actually lives. Exponentiating turns that 1 into \(e\), not the expression itself into 1.
Bottom line
The limit isn't 1 because you can't freeze the base at its limiting value while the exponent is still racing to infinity. \(1^\infty\) is indeterminate, and here the two competing effects balance exactly to produce \(e\).
From this definition, we derive the standard exponential limit:
The function approaches 1 as \(x \to 0\) from both directions. Note the asymmetric behavior: grows without bound as \(x \to \infty\), approaches 0 as \(x \to -\infty\).
The emerald-boxed limit above was stated without proof. Here we prove it directly from the definition of \(e\), and then use it to differentiate \(e^x\) from first principles.
Step 1: Start from the definition of the derivative.
For any function \(f(x)\), the derivative at \(x\) is
$$ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}. $$Taking \(f(x) = e^x\), this becomes
$$ \frac{d}{dx}(e^x) = \lim_{h \to 0} \frac{e^{x+h} - e^x}{h}. $$Step 2: Factor the numerator.
Using the exponent rule \(e^{x+h} = e^x \cdot e^h\):
$$ \frac{e^{x+h} - e^x}{h} = \frac{e^x e^h - e^x}{h} = \frac{e^x\,(e^h - 1)}{h}. $$So the derivative becomes
$$ \frac{d}{dx}(e^x) = \lim_{h \to 0} e^x \cdot \frac{e^h - 1}{h}. $$Step 3: Pull out the \(e^x\).
The limit variable is \(h\), and \(e^x\) does not depend on \(h\), so it behaves like a constant with respect to this limit:
$$ \frac{d}{dx}(e^x) = e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h}. $$All the work now reduces to the single limit
$$ L := \lim_{h \to 0} \frac{e^h - 1}{h}. $$Step 4: Show that \(\displaystyle\lim_{h \to 0} \frac{e^h - 1}{h} = 1\).
One standard way uses the definition of \(e\) as a limit together with a change of variables.
So we have established
$$ \lim_{h \to 0} \frac{e^h - 1}{h} = 1. $$(Alternatively, this same limit can be shown using the Squeeze Theorem or a power-series expansion of \(e^h\).)
Step 5: Plug this limit back in.
Returning to the derivative and using \(L = 1\):
$$ \frac{d}{dx}(e^x) = e^x \cdot \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \cdot 1 = e^x. $$This completes the limit-definition proof that the derivative of \(e^x\) is itself: \(\frac{d}{dx}(e^x) = e^x\). The standard limit \(\lim_{x \to 0}\frac{e^x-1}{x}=1\) is exactly what makes \(e^x\) the unique exponential whose slope equals its own height. Section 5 revisits this derivative connection for \(\sin x\) and \(\ln x\).
Derivation: We express \(a^x\) using the natural exponential function:
$$ a^x = e^{\ln(a^x)} = e^{x \ln a} $$Substituting into our limit:
$$ \lim_{x \to 0} \frac{a^x - 1}{x} = \lim_{x \to 0} \frac{e^{x \ln a} - 1}{x} $$Let \(u = x \ln a\). As \(x \to 0\), we have \(u \to 0\). Also, \(x = \frac{u}{\ln a}\), so:
$$ = \lim_{u \to 0} \frac{e^u - 1}{u/\ln a} = \lim_{u \to 0} \frac{e^u - 1}{u} \cdot \ln a = 1 \cdot \ln a = \ln a $$Special Cases:
These standard limits are not random. They resolve the 0/0 indeterminate form inherent in the definition of the derivative.
Recall the definition of the derivative: \(f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}\).
Example: Evaluate \(\lim_{x \to 0} \frac{\sin(3x)}{e^{2x} - 1}\).
We cannot substitute (0/0). We must algebraically manipulate the expression to use our standard limits. Divide numerator and denominator by x:
$$ \lim_{x \to 0} \frac{\frac{\sin(3x)}{x}}{\frac{e^{2x}-1}{x}} $$Apply the substitution property for limits (\(\lim_{x\to 0} \frac{\sin(kx)}{x} = k\)):
Result: \(\frac{3}{2}\)
Taylor series provide a powerful "power tool" way to view limits. If we replace functions with their polynomial approximations near 0:
Revisiting the Standard Limit:
$$ \lim_{x \to 0} \frac{\sin x}{x} \approx \lim_{x \to 0} \frac{x - x^3/6}{x} = \lim_{x \to 0} \left(1 - \frac{x^2}{6}\right) = 1 $$This demonstrates that as \(x \to 0\), the function \(\sin x\) is asymptotically equivalent to the linear term x.
Notice how the orange line (x) hugs the blue sine curve perfectly near x=0.
For limits yielding the indeterminate form 0/0 or \(\infty/\infty\), we can differentiate the numerator and denominator separately.
L'Hôpital's Rule (Complete Statement):
If \(\lim_{x \to a} f(x) = \lim_{x \to a} g(x) = 0\) (or both \(= \pm\infty\)), then:
$$ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} $$provided the following conditions are met:
Warning: L'Hôpital's Rule only works for \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\) forms. Other indeterminate forms like \(0 \cdot \infty\), \(\infty - \infty\), \(0^0\), \(1^\infty\), or \(\infty^0\) must first be algebraically converted to one of these forms.
Example: \(\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}\) (Form 0/0)
Apply L'Hôpital: \(\lim_{x \to 0} \frac{e^x - 1}{2x}\) (Still 0/0)
Apply L'Hôpital again: \(\lim_{x \to 0} \frac{e^x}{2} = \frac{1}{2}\).
In Multivariable Calculus, limits are harder because you can approach (0,0) from infinite directions. However, standard limits still apply if the argument goes to 0.
$$ \lim_{(x,y) \to (0,0)} \frac{\sin(xy)}{xy} = 1 $$Since as \((x,y) \to (0,0)\), the product \(u = xy \to 0\).
To prove a limit rigorously, we discard "approaches" and use inequalities. \(\lim_{x \to a} f(x) = L\) means: For every \(\epsilon \gt 0\) (error tolerance), there exists a \(\delta \gt 0\) (input proximity) such that:
$$ 0 \lt |x - a| \lt \delta \implies |f(x) - L| \lt \epsilon $$Proof Sketch for \(\lim_{x \to 3} (2x - 1) = 5\):
Audience Note: This section introduces Nonstandard Analysis, an alternative foundation for calculus developed in the 1960s. While not required for standard calculus courses, it provides historical context and appeals to those interested in mathematical logic or the philosophy of mathematics. Feel free to skip ahead to "Connections to Advanced Fields" if desired.
In the 1960s, Abraham Robinson rigorously justified the use of infinitesimals (numbers smaller than any real number but greater than 0), denoted by \(\varepsilon\) (epsilon) or dx.
Example: \(\lim_{x \to 0} \frac{\sin x}{x}\)
In the hyperreal system, if dx is infinitesimal:
$$ \frac{\sin(dx)}{dx} = 1 - \frac{dx^2}{3!} + \dots \approx 1 $$Taking the standard part: \(st(1 - \dots) = 1\).
| Type | Limit Formula | Condition |
|---|---|---|
| Power | \(\lim_{x \to a} x^n = a^n\) | \(n \gt 0\) |
| Trig | \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) | x in radians |
| Trig | \(\lim_{x \to 0} \frac{\tan x}{x} = 1\) | x in radians |
| Trig | \(\lim_{x \to 0} \frac{1 - \cos x}{x} = 0\) | |
| Trig | \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\) | |
| Log | \(\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1\) | |
| Exp | \(\lim_{x \to 0} \frac{e^x - 1}{x} = 1\) | |
| Euler | \(\lim_{n \to \infty} (1 + \frac{1}{n})^n = e\) | |
| General Exp | \(\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a\) | \(a \gt 0\) |
1. (Intermediate) Evaluate \(\lim_{x \to 0} \frac{1 - \cos(2x)}{x^2}\).
(Hint: Use the half-angle identity \(1-\cos(2x) = 2\sin^2(x)\))
2. (Advanced) Evaluate \(\lim_{x \to \infty} x \left( \ln(x+1) - \ln(x) \right)\).
(Hint: Combine logs, then substitute \(h = 1/x\))
3. (Expert) Using the Taylor expansion of \(\sin x\), prove that \(\lim_{x \to 0} \frac{\sin x - x}{x^3} = -\frac{1}{6}\).
Created for the dedicated student of Mathematics.