12.1 Finding limits: numerical and graphical approaches  (Page 6/10)

$\underset{x\to -2^{-}}{\lim }f(x)$

$\underset{x\to -2^{+}}{\lim }f(x)$

$\underset{x\to -2}{\lim }f(x)$

$f(\mathrm{-2})$

$\underset{x\to -1^{-}}{\lim }f(x)$

$\underset{x\to 1^{+}}{\lim }f(x)$

$\underset{x\to 1}{\lim }f(x)$

$f(1)$

$\underset{x\to 4^{-}}{\lim }f(x)$

$\underset{x\to 4^{+}}{\lim }f(x)$

$\underset{x\to 4}{\lim }f(x)$

$f(4)$

$\underset{x\to 0^{-}}{\lim }f(x)=2,\underset{x\to 0^{+}}{\lim }f(x)=–3,\underset{x\to 2}{\lim }f(x)=2,f(0)=4,f(2)=–1,f(–3)\text{ does not exist}.$

$\underset{x\to 2^{-}}{\lim }f(x)=0,\underset{x\to 2^{+}}{\lim }=–2,\underset{x\to 0}{\lim }f(x)=3,f(2)=5,f(0)$

$\underset{x\to 2^{-}}{\lim }f(x)=2,\underset{x\to 2^{+}}{\lim }f(x)=-3,\underset{x\to 0}{\lim }f(x)=5,f(0)=1,f(1)=0$

$\underset{x\to 3^{-}}{\lim }f(x)=0,\underset{x\to 3^{+}}{\lim }f(x)=5,\underset{x\to 5}{\lim }f(x)=0,f(5)=4,f(3)\text{ does not exist}.$

$\underset{x\to 4}{\lim }f(x)=6,\underset{x\to 6^{+}}{\lim }f(x)=-1,\underset{x\to 0}{\lim }f(x)=5,f(4)=6,f(2)=6$

$\underset{x\to -3}{\lim }f(x)=2,\underset{x\to 1^{+}}{\lim }f(x)=-2,\underset{x\to 3}{\lim }f(x)=–4,f(–3)=0,f(0)=0$

$\underset{x\to \pi }{\lim }f(x)={\pi }^2,\underset{x\to –\pi }{\lim }f(x)=\frac{\pi }{2},\underset{x\to 1^{–}}{\lim }f(x)=0,f(\pi )=\sqrt{2},f(0)\text{ does not exist}.$

$f(x)={\left(1+x\right)}^{\frac{1}{x}}$

$g(x)={\left(1+x\right)}^{\frac{2}{x}}$

$h(x)={\left(1+x\right)}^{\frac{3}{x}}$

$i(x)={\left(1+x\right)}^{\frac{4}{x}}$

$j(x)={\left(1+x\right)}^{\frac{5}{x}}$

Based on the pattern you observed in the exercises above, make a conjecture as to the limit of $f(x)={\left(1+x\right)}^{\frac{6}{x}},$ $g(x)={\left(1+x\right)}^{\frac{7}{x}},$ $\text{and }h(x)={\left(1+x\right)}^{\frac{n}{x}}.$

$(x)=\{\begin{array}{ll}\left|x\right|-1,\hfill & \text{if }x\ne 1\hfill \\ x^3,\hfill & \text{if }x=1\hfill \end{array}a=1$

$(x)=\{\begin{array}{ll}\frac{1}{x+1},\hfill & \text{if }x=-2\hfill \\ {(x+1)}^2,\hfill & \text{if }x\ne -2\hfill \end{array}a=-2$

$f(x)=\frac{x^2-4x}{16-x^2};a=4$

$f(x)=\frac{x^2-x-6}{x^2-9};a=3$

$f(x)=\frac{x^2-6x-7}{x^2– 7x};a=7$

$f(x)=\frac{x^2–1}{x^2–3x+2};a=1$

$f(x)=\frac{1-x^2}{x^2-3x+2};a=1$

$f(x)=\frac{10-10x^2}{x^2-3x+2};a=1$

$f(x)=\frac{x}{6x^2-5x-6};a=\frac{3}{2}$

$f(x)=\frac{x}{4x^2+4x+1};a=-\frac{1}{2}$

$f(x)=\frac{2}{x-4};a=4$

$\underset{x\to 0}{\lim }\frac{7\tan x}{3x}$

$\underset{x\to 4}{\lim }\frac{x^2}{x-4}$

$\underset{x\to 0}{\lim }\frac{2\sin x}{4\tan x}$

$\underset{x\to 0}{\lim }{e}^{e^{\frac{1}{x}}}$

$\underset{x\to 0}{\lim }{e}^{e^{-\frac{1}{x^2}}}$

$\underset{x\to 0}{\lim }\frac{\left|x\right|}{x}$

$\underset{x\to -1}{\lim }\frac{\left|x+1\right|}{x+1}$

$\underset{x\to 5}{\lim }\frac{\left|x-5\right|}{5-x}$

$\underset{x\to -1}{\lim }\frac{1}{{\left(x+1\right)}^2}$

$\underset{x\to 1}{\lim }\frac{1}{{\left(x-1\right)}^3}$

$\underset{x\to 0}{\lim }\frac{5}{1-e^{\frac{2}{x}}}$

Use numerical and graphical evidence to compare and contrast the limits of two functions whose formulas appear similar: $f(x)=\left|\frac{1-x}{x}\right|$ and $g(x)=\left|\frac{1+x}{x}\right|$ as $x$ approaches 0. Use a graphing utility, if possible, to determine the left- and right-hand limits of the functions $f\left(x\right)$ and $g\left(x\right)$ as $x$ approaches 0. If the functions have a limit as $x$ approaches 0, state it. If not, discuss why there is no limit.

According to the Theory of Relativity, the mass $m$ of a particle depends on its velocity $v$ . That is

where $m_o$ is the mass when the particle is at rest and $c$ is the speed of light. Find the limit of the mass, $m,$ as $v$ approaches $c^{-}.$

Allow the speed of light, $c,$ to be equal to 1.0. If the mass, $m,$ is 1, what occurs to $m$ as $v\to c?$ Using the values listed in [link] , make a conjecture as to what the mass is as $v$ approaches 1.00.