90 lines
3 KiB
Markdown
90 lines
3 KiB
Markdown
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# Limits
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If $f(x)$ is defined for all $x$ near a, except possibly at a itself, and if it can be ensured that $f(x)$ is as close to $L$ by taking $x$ close enough to $a$, but not equal to $a$. Then $f$ approaches the **limit** $L$ as $x$ approaches $a$:
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$$
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\lim_{x \to a} f(x) = L
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$$
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## One-sided limits
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If $f(x)$ is defined on some interval $(b,a)$ extending to the left of $x=a$, and if it can be ensured that $f(x)$ is as close to $L$ by taking $x$ to the left of $a$ and close enough to $a$, then $f(x) has **left limit** $L$ at $x=a$ and:
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$$
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\lim_{x \uparrow a} f(x) = L.
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$$
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If $f(x)$ is defined on some interval $(b,a)$ extending to the right of $x=a$ and if it can be ensured that $f(x)$ is as close to $L$ by taking $x$ to the right of $a$ and close enough to $a$, then $f(x) has **right limit** $L$ at $x=a$ and:
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$$
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\lim_{x \downarrow a} f(x) = L.
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$$
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## Limits at infinity
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If $f(x)$ is defined on an interval $(a,\infty)$ and if it can be ensured that $f(x)$ is as close to $L$ by taking $x$ large enough, then $f(x)$ **approaches the limit $L$ as $x$ approaches infinity** and
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$$
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\lim_{x \to \infty} f(x) = L
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$$
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## Limit rules
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If $\lim_{x \to a} f(x) = L$, $\lim_{x \to a} g(x) = M$, and $k$ is a constant then,
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* **Limit of a sum:** $\lim_{x \to a}[f(x) + g(x)] = L + M$.
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* **Limit of a difference:** $\lim_{x \to a}[f(x) - g(x)] = L - M$.
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* **Limit of a multiple:** $\lim_{x \to a}k f(x) = k L$.
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* **Limit of a product:** $\lim_{x \to a}f(x) g(x) = L M$.
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* **Limit of a quotient:** $\lim_{x \to a}\frac{f(x)}{g(x)} = \frac{L}{M}$, if $M \neq 0$.
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* **Limit of a power:** $\lim_{x \to a}[f(x)]^\frac{m}{n} = L^{\frac{m}{n}}$.
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## Formal definition of a limit
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The limit $\lim_{x \to a} f(x) = L$ means,
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$$
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\forall \varepsilon > 0, \exists \delta \space \mathrm{,s.t.,} \space
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\forall x \in \mathbb{R}, \space 0<|x-a|<\delta \implies |f(x) - L| < \varepsilon
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$$
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For one-sided, infinite and limits at infinity there are similar formal definitions.
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### Example
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Applying the formal definition of a limit for $\lim_{x \to 4}\sqrt{2x + 1}$
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* Given $\varepsilon > 0$
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* Choose $\delta = \frac{\varepsilon}{2}$
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* Suppose $0 < |x - 4| < \delta$
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* Check $|\sqrt{2x + 1} - 3|$
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$$
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\begin{array}{ll}
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|\sqrt{2x + 1} - 3| &= |\frac{(\sqrt{2x + 1} - 3)(\sqrt{2x + 1} + 3)}{\sqrt{2x + 1} + 3}|\\
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&= \frac{2|x - 4|}{\sqrt{2x + 1} + 3}\\
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&< 2|x-4|\\
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&< 2\delta = \varepsilon
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\end{array}
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$$
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## Squeeze Theorem
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Suppose that $f(x) \leq g(x) \leq h(x)$ holds for all $x$ in some open interval containing $a$, except possibly at $x=a$ itself. Suppose also that
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$$\lim_{x \to a} f(x) = \lim_{x \to a} h(x) = L.$$
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Then $\lim_{x \to a} g(x) = L$ also. Similar statements hold for left and right limits.
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### Example
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Applying squeeze theorem on $\lim_{x \to 0} x^2 \cos(\frac{1}{x})$.
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$$
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\begin{array}{ll}
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\forall x \neq 0\\
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-1 \leq \cos(\frac{1}{x}) \leq 1 \implies -x^2 \leq x^2 \cos(\frac{1}{x}) \leq x^2\\
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\mathrm{Since,} \space \lim_{x \to 0} x^2 = \lim_{x \to 0} -x^2 = 0\\
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\lim_{x \to 0} x^2 \cos(\frac{1}{x}) = 0
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\end{array}
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$$
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