Series

Author: John J Weber III, PhD Corresponding Textbook Sections:

• Section 11.2 – Series

Prerequisite Knowledge

Calculus II

Partial Fractions

Given ${\displaystyle \frac{mx+n}{a{x}^2+bx+c}}$$\displaystyle \frac{mx+n}{ax^2+bx+c}$, $a\ne 0$$a\ne 0$, both $m,n\ne 0$$m,n\ne 0$, and $a{x}^2+bx+c$$ax^2+bx+c$ is factorable, then

$\frac{mx+n}{a{x}^2+bx+c}=\frac{A}{px+q}+\frac{B}{sx+t}$$\frac{mx+n}{ax^2+bx+c}=\frac{A}{px+q}+\frac{B}{sx+t}$

where $a{x}^2+bx+c=(px+q)(sx+t)$$ax^2+bx+c=(px+q)(sx+t)$.

To solve for $A$$A$ and $B$$B$:

$(px+q)(sx+t)[\frac{mx+n}{a{x}^2+bx+c}=\frac{A}{px+q}+\frac{B}{sx+t}]$$(px+q)(sx+t)\left[\frac{mx+n}{ax^2+bx+c}=\frac{A}{px+q}+\frac{B}{sx+t}\right]$

$\Rightarrow mx+n=A(sx+t)+B(px+q)$$\Rightarrow mx+n=A(sx+t)+B(px+q)$

Use any algebraic method to solve for $A$$A$ and $B$$B$.

Calculus I

Limits

Infinite Limits

• If ${\displaystyle \underset{x\to a}{lim}f(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\infty$, i.e., $f(x)$$f(x)$ becomes large without bound, then the limit does not exist.
• If ${\displaystyle \underset{x\to a}{lim}f(x)=-\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=-\infty$, i.e., $f(x)$$f(x)$ becomes small without bound, then the limit does not exist.

NOTE: $\mathrm{\infty }$$\infty$ is NOT a real number.

Limits at Infinity

Let $f$$f$ be a function defined on some open interval $(a,\mathrm{\infty })$$(a,\infty)$. Then $\underset{x\to \mathrm{\infty }}{lim}f(x)=L$$\lim_{x\rightarrow\infty}{f(x)}=L$ means that as $x$$x$ becomes large without bound, the values of $f(x)$$f(x)$ become arbitrarily close to $L$$L$.

NOTE: Since $\mathrm{\infty }$$\infty$ is NOT a real number, then $f(\mathrm{\infty })$$f(\infty)$ is NOT meaningful.

Indeterminate Forms of Limits

Indeterminate Quotients

1. Type ${\displaystyle \frac{0}{0}}$$\displaystyle\frac{0}{0}$: If ${\displaystyle \underset{x\to a}{lim}f(x)=0}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=0$ and ${\displaystyle \underset{x\to a}{lim}g(x)=0}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=0$, then ${\displaystyle \underset{x\to a}{lim}\frac{f(x)}{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}$ may or may not exist.
2. Type ${\displaystyle \frac{\mathrm{\infty }}{\mathrm{\infty }}}$$\displaystyle\frac{\infty}{\infty}$: If ${\displaystyle \underset{x\to a}{lim}f(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\infty$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\infty$, then ${\displaystyle \underset{x\to a}{lim}\frac{f(x)}{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{\frac{f(x)}{g(x)}}$ may or may not exist.

Indeterminate Differences

3. Type ${\displaystyle \mathrm{\infty }-\mathrm{\infty }}$$\displaystyle\infty -\infty$: If ${\displaystyle \underset{x\to a}{lim}f(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\infty$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\infty$, then ${\displaystyle \underset{x\to a}{lim}[f(x)-g(x)]}$$\displaystyle\lim_{x\rightarrow a}{\left[f(x)-g(x)\right]}$ may or may not exist.

Indeterminate Products

4. Type ${\displaystyle 0\cdot \mathrm{\infty }}$$\displaystyle 0\cdot\infty$: If ${\displaystyle \underset{x\to a}{lim}f(x)=0}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=0$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\infty$, then ${\displaystyle \underset{x\to a}{lim}[f(x)g(x)]}$$\displaystyle\lim_{x\rightarrow a}{\left[f(x)g(x)\right]}$ may or may not exist.

Indeterminate Powers

5. Type ${\displaystyle 0^0}$$\displaystyle 0^0$: If ${\displaystyle \underset{x\to a}{lim}f(x)=0}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=0$ and ${\displaystyle \underset{x\to a}{lim}g(x)=0}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=0$, then ${\displaystyle \underset{x\to a}{lim}f(x{)}^{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{f(x)^{g(x)}}$ may or may not exist.
6. Type ${\displaystyle {\mathrm{\infty }}^0}$$\displaystyle \infty^0$: If ${\displaystyle \underset{x\to a}{lim}f(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\infty$ and ${\displaystyle \underset{x\to a}{lim}g(x)=0}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=0$, then ${\displaystyle \underset{x\to a}{lim}f(x{)}^{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{f(x)^{g(x)}}$ may or may not exist.
7. Type ${\displaystyle 1^{\mathrm{\infty }}}$$\displaystyle 1^{\infty}$: If ${\displaystyle \underset{x\to a}{lim}f(x)=1}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=1$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\infty$, then ${\displaystyle \underset{x\to a}{lim}f(x{)}^{g(x)}}$$\displaystyle\lim_{x\rightarrow a}{f(x)^{g(x)}}$ may or may not exist.

l'Hôpital's Rule

NOTE: l'Hôpital's Rule is valid only for indeterminate quotients.

Definition

Suppose $f$$f$ and $g$$g$ are differentiable and $g^{\prime }(x)\ne 0$$g^{\,\prime}(x)\ne 0$ on some open interval that contains $a$$a$. Suppose that

• ${\displaystyle \underset{x\to a}{lim}f(x)=0}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=0$ and ${\displaystyle \underset{x\to a}{lim}g(x)=0}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=0$; or
• ${\displaystyle \underset{x\to a}{lim}f(x)=\pm \mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{f(x)}=\pm\infty$ and ${\displaystyle \underset{x\to a}{lim}g(x)=\pm \mathrm{\infty }}$$\displaystyle\lim_{x\rightarrow a}{g(x)}=\pm\infty$

then $\underset{x\to a}{lim}{\displaystyle \frac{f(x)}{g(x)}}=\underset{x\to a}{lim}{\displaystyle \frac{f^{\prime }(x)}{g^{\prime }(x)}}$$\lim_{x\rightarrow a}{\dfrac{f(x)}{g(x)}}=\lim_{x\rightarrow a}{\dfrac{f^{\,\prime}(x)}{g^{\,\prime}(x)}}$

Derivatives

• ${\displaystyle \frac{d}{dx}c=0}$$\displaystyle \frac{d}{dx}c=0$
• ${\displaystyle \frac{d}{dx}[cf(x)]=c{f}^{\prime }(x)}$$\displaystyle \frac{d}{dx}\left[cf(x)\right]=cf^{\,\prime}(x)$
• ${\displaystyle \frac{d}{dx}[f(x)+g(x)]=f^{\prime }(x)+g^{\prime }(x)}$$\displaystyle \frac{d}{dx}\left[f(x)+g(x)\right]=f^{\,\prime}(x)+g^{\,\prime}(x)$
• ${\displaystyle \frac{d}{dx}[f(x)g(x)]=f^{\prime }(x)g(x)+f(x)g^{\prime }(x)}$$\displaystyle \frac{d}{dx}\left[f(x)g(x)\right]=f^{\,\prime}(x)g(x)+f(x)g^{\,\prime}(x)$
• ${\displaystyle \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{{[g(x)]}^2}}$$\displaystyle \frac{d}{dx}\frac{f(x)}{g(x)}=\frac{f^{\,\prime}(x)g(x)-f(x)g^{\,\prime}(x)}{\left[g(x)\right]^2}$
• ${\displaystyle \frac{d}{dx}f(g(x))=f^{\prime }(g(x))g^{\prime }(x)}$$\displaystyle \frac{d}{dx}f(g(x))=f^{\,\prime}(g(x))g^{\,\prime}(x)$
• ${\displaystyle \frac{d}{dx}{x}^n=n{x}^{n-1}}$$\displaystyle \frac{d}{dx}x^n=nx^{n-1}$
• ${\displaystyle \frac{d}{dx}{e}^x=e^x}$$\displaystyle \frac{d}{dx}e^x=e^x$
• ${\displaystyle \frac{d}{dx}{a}^x=\ln (x)a^x}$$\displaystyle \frac{d}{dx}a^x=\ln{(x)}a^x$, $a>0$$a>0$, $a\ne 1$$a\ne 1$
• ${\displaystyle \frac{d}{dx}\ln (x)=\frac{1}{x}}$$\displaystyle \frac{d}{dx}\ln{(x)}=\frac{1}{x}$
• ${\displaystyle \frac{d}{dx}{\log }_b(x)=\frac{1}{\ln (b)x}}$$\displaystyle \frac{d}{dx}\log_b{(x)}=\frac{1}{\ln{(b)}x}$, $b>0$$b>0$, $b\ne 1$$b\ne 1$

College Algebra

Absolute Value Inequality

$|ax+b|<c$$\left|ax+b\right|<c$

Procedure

1. Rewrite as a compound inequality: $-c<ax+b<c$$-c<ax+b<c$
2. Subtract the constant $b$$b$ from all parts of compound inequality: $-c-b<ax<c-b$$-c-b<ax<c-b$
3. Divide all parts of compound inequality by coefficient of $x$$x$: ${\displaystyle \frac{-c-b}{a}<x<\frac{c-b}{a}}$$\displaystyle \frac{-c-b}{a}<x<\frac{c-b}{a}$
4. Write answer in interval notation: ${\displaystyle (\frac{-c-b}{a},\frac{c-b}{a})}$$\displaystyle \left(\frac{-c-b}{a},\frac{c-b}{a}\right)$

$|ax+b|\le c$$\left|ax+b\right|\le c$

Procedure

1. Rewrite as a compound inequality: $-c\le ax+b\le c$$-c\le ax+b\le c$
2. Subtract the constant $b$$b$ from all parts of compound inequality: $-c-b\le ax\le c-b$$-c-b\le ax\le c-b$
3. Divide all parts of compound inequality by coefficient of $x$$x$: ${\displaystyle \frac{-c-b}{a}\le x\le \frac{c-b}{a}}$$\displaystyle \frac{-c-b}{a}\le x\le\frac{c-b}{a}$
4. Write answer in interval notation: ${\displaystyle [\frac{-c-b}{a},\frac{c-b}{a}]}$$\displaystyle \left[\frac{-c-b}{a},\frac{c-b}{a}\right]$

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Last Modified: Tuesday, 17 November 2020 5:50 EDT