\lim_{x \rightarrow a} f(x) = L means that f(x) gets arbitrarily close to L as x gets arbitrarily close to a.
Following are three variables f, g, and c:
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}100 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}100 |
f and g refer to functions, and c refers to a constant.
You can change them to any other values you'd like as you work through the examples.
1. \lim_{x \rightarrow c}(f(x) + g(x)) = \lim_{x \rightarrow c}f(x) + \lim_{x \rightarrow c}g(x)
The limit of a sum is the sum of the limits.
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} + 2 \, x - 2 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} + 2 \, x - 2 |
\lim_{x \rightarrow c} f(x), \lim_{x \rightarrow c} g(x):
\newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}9999 \newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}9999 |
\lim_{x \rightarrow c}(f + g)(x)
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} + 2 \, x - 2 \newcommand{\Bold}[1]{\mathbf{#1}}100 \newcommand{\Bold}[1]{\mathbf{#1}}10198 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} + 2 \, x - 2 \newcommand{\Bold}[1]{\mathbf{#1}}100 \newcommand{\Bold}[1]{\mathbf{#1}}10198 |
\lim_{x=c}f(x) + \lim_{x=c}g(x)
\newcommand{\Bold}[1]{\mathbf{#1}}10198
\newcommand{\Bold}[1]{\mathbf{#1}}10198
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2. \lim_{x \rightarrow c}(f(x) - g(x)) = \lim_{x \rightarrow c}f(x) - \lim_{x \rightarrow c}g(x)
The limit of a difference is the difference of the limits.
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ -x^{2} + 2 \, x \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ -x^{2} + 2 \, x |
\lim_{x \rightarrow c} f(x), \lim_{x \rightarrow c} g(x):
\newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}9999 \newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}9999 |
\lim_{x \rightarrow c}(f - g)(x)
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ -x^{2} + 2 \, x \newcommand{\Bold}[1]{\mathbf{#1}}-9800 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ -x^{2} + 2 \, x \newcommand{\Bold}[1]{\mathbf{#1}}-9800 |
\lim_{x=c}f(x) - \lim_{x=c}g(x)
\newcommand{\Bold}[1]{\mathbf{#1}}-9800
\newcommand{\Bold}[1]{\mathbf{#1}}-9800
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3. \lim_{x \rightarrow c}(f(x) \cdot g(x)) = \lim_{x \rightarrow c}f(x) \cdot \lim_{x \rightarrow c}g(x)
The limit of a product is the product of the limits.
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ {\left(2 \, x - 1\right)} {\left(x^{2} - 1\right)} \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ {\left(2 \, x - 1\right)} {\left(x^{2} - 1\right)} |
\lim_{x \rightarrow c} f(x), \lim_{x \rightarrow c} g(x):
\newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}9999 \newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}9999 |
\lim_{x \rightarrow c}(f \cdot g)(x)
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ {\left(2 \, x - 1\right)} {\left(x^{2} - 1\right)} \newcommand{\Bold}[1]{\mathbf{#1}}1989801 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ {\left(2 \, x - 1\right)} {\left(x^{2} - 1\right)} \newcommand{\Bold}[1]{\mathbf{#1}}1989801 |
\lim_{x=c}f(x) \cdot \lim_{x=c}g(x)
\newcommand{\Bold}[1]{\mathbf{#1}}1989801
\newcommand{\Bold}[1]{\mathbf{#1}}1989801
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4. \lim_{x \rightarrow c}(k \cdot g(x)) = k \cdot \lim_{x \rightarrow c}g(x)
The limit of a scaling is the scaling of the limit.
Here we're going to introduce another variable k to represent a scalar.
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\lim_{x \rightarrow c} g(x)
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}9999 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}9999 |
\lim_{x \rightarrow c}k \cdot g(x)
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 4 \, x^{2} - 4 \newcommand{\Bold}[1]{\mathbf{#1}}39996 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 4 \, x^{2} - 4 \newcommand{\Bold}[1]{\mathbf{#1}}39996 |
k\lim_{x \rightarrow c}g(x)
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5. \lim_{x \rightarrow c}\frac{f(x)}{g(x)} = \frac{\lim_{x \rightarrow c}f(x)}{\lim_{x \rightarrow c}g(x)}
The limit of a ratio is the ratio of the limits.
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ \frac{2 \, x - 1}{x^{2} - 1} \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ x^{2} - 1 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ \frac{2 \, x - 1}{x^{2} - 1} |
\lim_{x \rightarrow c} f(x), \lim_{x \rightarrow c} g(x):
\newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}9999 \newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}9999 |
\lim_{x \rightarrow c}(\frac{f}{g})(x)
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ \frac{2 \, x - 1}{x^{2} - 1} \newcommand{\Bold}[1]{\mathbf{#1}}\frac{199}{9999} \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ \frac{2 \, x - 1}{x^{2} - 1} \newcommand{\Bold}[1]{\mathbf{#1}}\frac{199}{9999} |
\frac{\lim_{x=c}f(x)}{ \lim_{x=c}g(x)}
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{199}{9999}
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{199}{9999}
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6. lim_{x \rightarrow c} (f(x))^n = (lim_{x \rightarrow c} f(x))^n
The limit of a power is the power of the limit.
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\lim_{x \rightarrow c} (f(x))^n
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ {\left(2 \, x - 1\right)}^{5} \newcommand{\Bold}[1]{\mathbf{#1}}312079600999 \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ {\left(2 \, x - 1\right)}^{5} \newcommand{\Bold}[1]{\mathbf{#1}}312079600999 |
(\lim_{x \rightarrow c}f(x))^n
\newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}312079600999 \newcommand{\Bold}[1]{\mathbf{#1}}199 \newcommand{\Bold}[1]{\mathbf{#1}}312079600999 |
7. \lim_{x \rightarrow c} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \rightarrow c} f(x)}
The limit of a root is the root of the limit.
\lim_{x \rightarrow c} \sqrt[n]{f(x)}
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}5 \newcommand{\Bold}[1]{\mathbf{#1}}199^{\frac{1}{5}} \newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ 2 \, x - 1 \newcommand{\Bold}[1]{\mathbf{#1}}5 \newcommand{\Bold}[1]{\mathbf{#1}}199^{\frac{1}{5}} |
\sqrt[n]{\lim_{x \rightarrow c} f(x)}
\newcommand{\Bold}[1]{\mathbf{#1}}199^{\frac{1}{5}}
\newcommand{\Bold}[1]{\mathbf{#1}}199^{\frac{1}{5}}
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lim_{x \rightarrow \infty} f(x) = L means that f(x) gets arbitrarily close to L as x gets arbitrarily large.
lim_{x \rightarrow -\infty} f(x) = L means that f(x) gets arbitrarily close to L as - x gets arbitrarily large.
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ \frac{{\left(x + 1\right)}^{2}}{x^{2} - 3}
\newcommand{\Bold}[1]{\mathbf{#1}}x \ {\mapsto}\ \frac{{\left(x + 1\right)}^{2}}{x^{2} - 3}
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\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1002001}{999997}
\newcommand{\Bold}[1]{\mathbf{#1}}\frac{1002001}{999997}
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\newcommand{\Bold}[1]{\mathbf{#1}}1
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\newcommand{\Bold}[1]{\mathbf{#1}}1
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\newcommand{\Bold}[1]{\mathbf{#1}}1
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This is a useful limit in the study of Calculus.
Here is an elegant proof of it:
Imagine an n-sided regular polygon inscribed in a circle of radius r.
The polygon is composed of \frac{2\pi}{x} isosceles triangles where the vertex angle = x.
The area of one of these triangles is \frac{1}{2} r^2 \sin{x}, and
the area of all of these triangles is \frac{2\pi}{x} \cdot \frac{1}{2} r^2 \sin{x}.
As the angle x decreases, the number of triangles will increase,
and the sum of their areas will get arbitrarily close to the area of the circle.
\frac{2\pi}{x} \cdot \frac{1}{2} r^2 \sin{x} simplifies quite nicely to \pi r^2 \frac{\sin x}{x}.
Therefore, \lim_{x \rightarrow 0} \pi r^2 \frac{\sin x}{x} = \pi r^2,
and \pi r^2 \cdot \lim_{x \rightarrow 0}\frac{\sin x}{x} = \pi r^2,
proving that \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1.
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