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This is a part of [[Calculus 1: course]]. ==New Limits from Old== Follows the idea of making ''[[new functions from old]]''. How? Algebraic operations $ +, -, \cdot, \div $ '''Example:''' $\underbrace{e^{x}}_{\text{continuous}} + \underbrace{x}_{\text{continuous}}$. Is it continuous? Consider: $$ \lim_{x \to 0} e^{x} = 1, \;\;\; \lim_{x \to 0} x = 0$$ Thus $$ \lim_{x \to 0} (e^{x} + x) = 1 + 0 $$ '''Rule:''' ''The limit of the sum is equal to the sum of the limits.'' Sidenote: $$ \sqrt{3 + 2} \neq \sqrt{3} + \sqrt{2} $$ Consider how the rule works numerically. {| class="wikitable" style="text-align:center" border="1" |- ! $x$ !! $e^{x}$ !! $x$ !! $e^{x} + x $ |- | $1$ || $e$ || $1$ || $ e + 1 $ |- | $\frac{1}{2}$ || $\sqrt{e}$ || $\frac{1}{2}$ || $\sqrt{e} + \frac{1}{2}$ |- | $\frac{1}{3}$ || $\sqrt[3]{e}$ || $\frac{1}{3}$ || $\sqrt[3]{e} + \frac{1}{3}$ |- | $\vdots$ || $\vdots$ || $\vdots$ || $\vdots$ |- | $0$ || $1$ || $0$ || $1 + 0 = 1$ |} (Also, ''difference, product and quotient'' lead to a similar table.) '''[[Sum Rule]]:''' Suppose $\lim\limits_{x \to a} f(x) = L$, $\lim\limits_{x \to a} g(x)= M$ (this means that we assume both exist!). Then $\lim\limits_{x \to a} (f(x) + g(x)) = L + M. $ Which means that $\lim\limits_{x \to a} (f(x) + g(x))$ also exists. Similarly, $$ \begin{aligned} \lim_{x \to a} (f(x) - g(x)) & = L - M \\ \lim_{x \to a} (f(x) \times g(x)) & = L \times M \\ \lim_{x \to a} (f(x) \div g(x)) & = L \div M \end{aligned} $$ (they all exist.) '''Example:''' Does $\lim\limits_{x \to a} (f(x)+g(x))$ exist, even though $\lim\limits_{x \to a} f(x)$ and $\lim\limits_{x \to a} g(x)$ do not? Can the addition of $f + g$ cancel the irregular behavior? ''Yes.'' Solution: What if $g = -f$? Then $f + g = 0$, so $\lim\limits_{x \to a} (f + g) = 0$. It exists. For example: Take $f(x) = \dfrac{1}{x}; a = 0$ The $\lim\limits_{x \to 0} \dfrac{1}{x}$ does not exist. Neither does, $\lim\limits_{x \to 0} \left(-\dfrac{1}{x}\right)$. But $$ \lim_{x \to 0} \left( \dfrac{1}{x} + \left(-\dfrac{1}{x} \right) \right) = \lim_{x \to 0} 0 = 0 $$ '''Theorem:''' If $\lim\limits_{x \to a} f(x)$ exist but $lim_{x \to a} g(x)$ does not, then $\lim\limits_{x \to a} (f(x) +g(x))$ does not exist. ''Prove it!'' '''Review Exercise''' Plot the graph of: $$ f(x) = \begin{cases} 2 - x & \text{ if } x < -1 \\ x & \text{ if } -1 < x < 1 \\ (x - 1)^{2} & \text{ if } x >1 \end{cases} $$ [[Image:Page 20.png]] $$ \begin{aligned} (1-1)^{2} & = 0 \\ (2-1)^{2} & = 1 \\ (3-1)^{2} & = 4 \end{aligned} $$ $\lim\limits_{x \to a} f(x)$ exists for all $a$ except $a = -1,1$ $$ \begin{aligned} \lim_{x \to -1^{-}} f(x) & = 3 \\ \lim_{x \to -1^{+}} f(x) & = -1 \\ \lim_{x \to 1^{-}} f(x) & = 1 \\ \lim_{x \to 1{+}} f(x) & = 0 \end{aligned} $$ The problem is about interpreting the graph in terms of limits. Now, the other way around. '''Exercise.''' From limits to graphs. Given $$\begin{alignat}{3} \lim_{x \to 0^{-}} f(x) & = 1, & \quad \lim_{x \to 0^{+}} f(x) &= -1, & \quad f(0) & \text{undefined}, \\ \lim_{x \to 2^{-}} f(x) &= 0, &\quad \lim_{x \to 2^{+}} f(x) & = 1, &\quad f(2) &= 1 \end{alignat} $$ As $x \to 0^{-}, y \to 1$<br/> As $x \to 0^{+}, y \to -1$<br/> As $x\to 2^{-}, y \to 0$<br/> As $x\to 2^{+}, y\to 1$<br/> Just concentrate on the behavior of $f$ close to these prints. Plot the graph of a possible $f$. [[Image:Page 21.png]] ==Algebraic Rules of Limits== How algebraic operations affect limits? Assume that $\lim\limits_{x \to a} f(x)$ and $\lim\limits_{x \to a} g(x)$ both exist. Then $$ \begin{aligned} \lim_{x \to a} f(x) & = L \\ \lim_{x \to a} g(x) & = M \end{aligned} $$ Then, it is easy to conjecture that the limits on the left exist, first, and equal to the numbers on the right: $$ \begin{alignat}{3} \lim_{x \to a} (f(x) + g(x)) & = L + M & \quad &\text{ Sum} \\ \lim_{x \to a} (f(x) - g(x)) & = L - M & \quad &\text{Difference} \\ \lim_{x \to a} (cf(x)) & = cL & \quad &\text{ Constant} \\ \lim_{x \to a} (f(x)\cdot g(x)) & = L\cdot M & \quad &\text{ Product} \\ \lim_{x \to a} \left(\dfrac{f(x)}{g(x)}\right) & = \dfrac{L}{M} & \quad &\text{ Quotient, provided }M \ne 0 \\ \end{alignat} $$ In the last case, $g(x)$ can be 0 for some $x$. '''Exercise:''' $f(x) = 1, g(x) = x, g(0) = 0$ Given a new function: $h(x) = \dfrac{f(x)}{g(x)}$, What is the domain of $h$? They are the values of $x$ for which $h(x)$ makes sense. If $g(0) = 0$ then 0 is not the domain of $h(x) = f(x)/g(x)$. Compute these limits: $\lim_{x \to 0} \dfrac{f(x)}{g(x)} = \lim_{x \to 0}\dfrac{1}{x}$ does not exist, but $\lim_{x\to 1} \dfrac{f(x)}{g(x)} = 1$ does as $M = \lim_{x \to 1} x = 1 \neq 0$. '''Example:''' $\lim_{x \to 2} 3 = 3$. Indeed, as $x\to 2, 3\to 3$ (where else?) as $y = 3$ does not go anywhere. '''Example:''' Justify $\lim_{x \to 2} {(x)}=2$ Trivial. As $x \to 2$ as $x \to 2$ (where else?). '''Example:''' $\lim{x \to 2} x^{2} = 2^{2} = 4$. Justify : $f(x) = x^2$ is [[continuous]] at 2! '''Very often:''' "$x \to 2$" is translated as $x = 2$, i/e., we evaluate the limit by ''[[substitution]]''. ==Substitution Property== Recall: If $f$ is ''[[continuous]]'' at $x = a$, then $\lim_{x \to a} f(x) = f(a)$. This is the definition of ''continuity''. It's also called the ''direct substitution rule''. But what about $\lim_{x \to 1} e^{x^{2}}$? What if don't know (yet) that the function is continuous? There is a better rule for this. '''Substitution Rule.''' $$\lim_{x \to a} g(f(x)) = g(\lim_{x \to a} f(x)),$$ if $g$ is continuous at $y = \lim_{x \to a} f(x)$ (we assume that this limit exists!). '''Example:''' Apply it. $\lim_{x \to 1} e^{x^{2}} = e^{\lim_{x \to 1} x^{2}} = e^{1^{2}} = e.$ This works because $e^{y}$ is continuous at $y = 1$. The property is about [[composition]]s of functions: $$ \xrightarrow[x]{} x^{2} \xrightarrow[y]{} e \xrightarrow[z]{} $$ '''Corollary''' If $f$ and $g$ are both continuous then $$\lim_{x \to a} g(f(x)) = g(f(a)).$$ '''Example''' $e^{y}$ and $x^{2}$ are continuous (everywhere). Then $$\lim_{x \to 1} e^{x^{2}} = e^{1^{2}} = e.$$ '''Example''' Justify this: $ \lim_{x \to 1} \sin \sqrt{x} = \sin \sqrt{1} = \sin 1.$ The ''substitution property'' applicable because $\sin y$ is continuous at $y = 1$ and $\sqrt{x}$ is continuous at 1. Compare to $\lim\limits_{x \to 0} \sin \sqrt{x}$ which is undefined as is $\lim\limits_{x \to 0^{+}} \sin \sqrt{x}$ Corollary does not apply here because $\sqrt{x}$ is not continuous at 0, as we will see later. '''Properties:''' (just choosing a specific $g$ in the substitution property) # $\lim\limits_{x \to a} [f(x)]^{n} = [\lim\limits_{x \to a}f(x)]^{n}$, chose $g(y) = y^{n}$ here. # $\lim\limits_{x \to a} \sqrt{f(x)} = \sqrt{\lim\limits_{x \to a} f(x)}$, if the right limit exist and is positive. ==What if we end up with 0/0?== '''Example:''' Recall that $\lim\limits_{x \to 1} \dfrac{(x-1)^{2}}{x-1}$ or $\dfrac{0}{0}$ Sidenote: Informally, $\dfrac{1}{0} = \infty$. Observe that the numerator is non-zero here... This expression, 0/0, is called and ''[[indeterminate]]'', what does it mean? ''We make no conclusions!'' And we try something else, which usually is algebra. $$ \lim \overset{\text{cancel}}{=} \lim_{x \to 1} (x - 1) \overset{\text{substitute}}{=} 1 - 1 = 0$$ '''Example:''' What should $\ast$ be in the limit $$\lim\limits_{x \to 1} \dfrac{\ast}{x-1} = 2$$ so that it makes sense? Try this: $$ \begin{aligned} \lim_{x \to 2} \dfrac{(x-1)\cdot2}{x - 1} & = \lim_{x \to 2} 2 \\ &= 2 \end{aligned} $$ What if we replace 2 with 17? Compute... In fact, this indeterminate expression, $\tfrac{0}{0}$, could be any number. Further, can it be $\infty$? Yes: $$\dfrac{0}{0} \gets \lim_{x \to 0} \dfrac{x - 1}{(x - 1)^{2}} = \lim_{x \to 1}\dfrac{1}{x - 1} = \infty ( = +\infty \text{ or } -\infty) $$ '''Example:''' $\lim\limits_{x \to 1} \dfrac{x^{2} -1}{x - 1} \rightarrow \dfrac{0}{0}$ If we mindlessly substitute, we get 0/0... What does it mean? It means '''STOP!''' Erase everything and do algebra. $$\begin{aligned} \lim\limits_{x \to 1} \dfrac{x^{2} -1}{x - 1} & = \lim_{x \to 1} \dfrac{(x - 1)(x+1)}{x-1} \\ & = \lim_{x \to 1} (x+1) \\ & = 2 \end{aligned} $$ '''Example:''' $\lim\limits_{x \to 0} \dfrac{\sqrt{x^{2} + 9} - 3}{x^{2}}$ Plug in: $\tfrac{\sqrt{0 + 9} - 3}{0} = \tfrac{0}{0}$. Not good. Then do algebra, the purpose? To cancel the denominator. '''Trick:''' Multiply by the "conjugate": $\sqrt{x^{2} + 9} + 3$. The goal is to ''"rationalize"'' the numerator. It works because of the formula: $$ (a-b)(a+b) = a^{2} - b^{2} $$ Thus $$ \begin{aligned} \lim_{x \to 0} \dfrac{\sqrt{x^{2} + 9} - 3}{x^{2}} & = \lim_{x \to 0} \dfrac{(\sqrt{x^{2} + 9} - 3)\cdot (\sqrt{x^{2} + 9} + 3)}{x^{2}\cdot (\sqrt{x^{2} + 9} + 3)} \\ & = \lim_{x \to 0} \dfrac{(x^{2} + 9) - 3^{2}}{x^{2}(\sqrt{x^{2} + 9} + 3)} \\ & = \lim_{x \to 0} \dfrac{x^{2}}{x^{2}(\sqrt{x^{2} + 9} + 3)} \\ & = \lim_{x \to 0} \dfrac{1}{\sqrt{x^{2} + 9} + 3} \\ & = \dfrac{1}{\sqrt{0 + 9} + 3} = \dfrac{1}{6} \end{aligned} $$ Still $\dfrac{0}{0}$, but not after canceling. QR applies because the limit in the denominator exists and is not 0. ==How to apply the rules== Be careful about applying the rules. For example, if you apply the sum rule: $$ \lim_{x \to a} (f(x) +g(x)) \overset{\text{SR}}{=} \lim_{x \to a} (f(x) + \lim_{x \to a} (g(x),$$ wrong, unless both limits exist. Also, "[[DNE]]" is not a number: $ \text{DNE} + 5$ is meaningless. Rule: ''Don't do algebra with "DNE".'' '''Possible outcomes of limits:''' $$ \lim = \begin{cases} L, \text{ number } & \rightarrow \text{ algebra } \\ \infty, +\infty, -\infty & \rightarrow \text{ some algebra: } \infty + \infty = \infty \\ \text{DNE} & \rightarrow \text{ no algebra} \end{cases}$$ It;s similar to what we observe when we deal with functions (no limits). Recall: Functions : [[Image:Page 31-a.png]] Sometimes : [[Image:Page 31-b.png]] Limits are similar : [[Image:Page 31-b.png]] There may be nothing here as the output. In particular, $$\left.\begin{aligned} \lim_{x \to a} f(x) &- \lim_{x \to a} f(x) \\ & \dfrac{\lim\limits_{x \to a} f(x)}{\lim\limits_{x \to a} f(x)} \end{aligned} \right\} \qquad \text{Do nothing}$$ Unless you know the limits exist. ==Review Exercise== $$\lim_{x \to -3^{-}} \dfrac{x+2}{x+3} = ?$$ Numerically: $$\begin{aligned} x & = 4, -3.5, -3.1, -3.01, \ldots \\ y & = \ldots, \ldots, \ldots, \ldots, \ldots, \to +\infty \end{aligned} $$ Incomplete... (same problem with plotting) '''Solution:''' (algebraical) $$ \begin{aligned} \lim_{x \to -3^{-}} (x + 2) & = -3 + 2 = -1 \quad \text{Since } x+2 \text{ is continuous.} \\ \lim_{x \to -3^{-}} (x + 3) & = -3 + 3 = 0 \quad \text{Since } x+3 \text{ is continuous.} \end{aligned} $$ '''Observe:''' Since limit of domain is 0, the [[quotient rule of limits|QR]] (above) does not apply. But, $\lim$ of the numerator is $0$! Therefore the limit of the quotient is [[infinite]]. '''Question:''' Which infinity? $$ \lim = +\infty \text { or } \lim = -\infty $$ $x \to -3^{-}$ means $x < -3$ and close to -3. If $x$ is like this, what is the sign of $f(x) = \dfrac{x+2}{x+3}$? If $$ \begin{aligned} x + 2 & < 0 \\ x + 3 & < 0 \end{aligned} $$ So $f(x) >0$ '''Answer:''' $\lim\limits_{x \to -3^{-}} \dfrac{x+2}{x+3} = +\infty$ Continue to [[Limits: part 3]]. [[category:calculus]]

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