2.5 The precise definition of a limit  (Page 6/8)

$\underset{x\to a}{\text{lim}}f\left(x\right)=N$

$\underset{t\to b}{\text{lim}}g\left(t\right)=M$

$\underset{x\to c}{\text{lim}}h\left(x\right)=L$

$\underset{x\to a}{\text{lim}}\phi \left(x\right)=A$

If $0<\left|x-2\right|<\delta ,$ then $\left|f\left(x\right)-2\right|<1.$

If $0<\left|x-2\right|<\delta ,$ then $\left|f\left(x\right)-2\right|<0.5.$

If $0<\left|x-3\right|<\delta ,$ then $\left|f\left(x\right)+1\right|<1.$

If $0<\left|x-3\right|<\delta ,$ then $\left|f\left(x\right)+1\right|<2.$

$\epsilon =1.5$

$\epsilon =3$

$\left|\text{sin}\left(2x\right)-\frac{1}{2}\right|<0.1,$ whenever $\left|x-\frac{\pi }{12}\right|<\delta$

$\left|\sqrt{x-4}-2\right|<0.1,\text{whenever}\left|x-8\right|<\delta$

$\underset{x\to 2}{\text{lim}}\left(5x+8\right)=18$

$\underset{x\to 3}{\text{lim}}\frac{x^2-9}{x-3}=6$

$\underset{x\to 2}{\text{lim}}\frac{2x^2-3x-2}{x-2}=5$

$\underset{x\to 0}{\text{lim}}{x}^4=0$

$\underset{x\to 2}{\text{lim}}(x^2+2x)=8$

$\underset{x\to 5^{-}}{\text{lim}}\sqrt{5-x}=0$

$\underset{x\to 0^{+}}{\text{lim}}f\left(x\right)=\mathrm{-2},\text{where}f\left(x\right)=\{\begin{array}{c}8x-3,\text{if}x<0\\ 4x-2,\text{if}x\ge 0\end{array}.$

$\underset{x\to 1^{-}}{\text{lim}}f\left(x\right)=3,\text{where}f\left(x\right)=\{\begin{array}{c}5x-2,\text{if}x<1\\ 7x-1,\text{if}x\ge 1\end{array}.$

$\underset{x\to 0}{\text{lim}}\frac{1}{x^2}=\infty$

$\underset{x\to \mathrm{-1}}{\text{lim}}\frac{3}{{\left(x+1\right)}^2}=\infty$

$\underset{x\to 2}{\text{lim}}-\frac{1}{{\left(x-2\right)}^2}=\text{−}\infty$

An engineer is using a machine to cut a flat square of Aerogel of area 144 cm ² . If there is a maximum error tolerance in the area of 8 cm ² , how accurately must the engineer cut on the side, assuming all sides have the same length? How do these numbers relate to $\delta ,$ ε , a , and L ?

Use the precise definition of limit to prove that the following limit does not exist: $\underset{x\to 1}{\text{lim}}\frac{\left|x-1\right|}{x-1}.$

Using precise definitions of limits, prove that $\underset{x\to 0}{\text{lim}}f\left(x\right)$ does not exist, given that $f\left(x\right)$ is the ceiling function. ( Hint : Try any $\delta <1\text{.)}$

Using precise definitions of limits, prove that $\underset{x\to 0}{\text{lim}}f\left(x\right)$ does not exist: $f\left(x\right)=\{\begin{array}{l}1\text{if}x\text{is rational}\\ 0\text{if}x\text{is irrational}\end{array}.$ ( Hint : Think about how you can always choose a rational number $0<r<d,$ but $\left|f\left(r\right)-0\right|=1\text{.)}$

Using precise definitions of limits, determine $\underset{x\to 0}{\text{lim}}f\left(x\right)$ for $f\left(x\right)=\{\begin{array}{l}x\text{if}x\text{is rational}\\ 0\text{if}x\text{is irrational}\end{array}.$ ( Hint : Break into two cases, x rational and x irrational.)

Using the function from the previous exercise, use the precise definition of limits to show that $\underset{x\to a}{\text{lim}}f\left(x\right)$ does not exist for $a\ne 0.$

$\underset{x\to a}{\text{lim}}\left(f\left(x\right)-g\left(x\right)\right)=L-M$

$\underset{x\to a}{\text{lim}}\left[cf\left(x\right)\right]=cL$ for any real constant c ( Hint : Consider two cases: $c=0$ and $c\ne 0\text{.)}$

$\underset{x\to a}{\text{lim}}\left[f\left(x\right)g\left(x\right)\right]=LM.$ ( Hint : $\left|f\left(x\right)g\left(x\right)-LM\right|=$ $\left|f\left(x\right)g\left(x\right)-f\left(x\right)M+f\left(x\right)M-LM\right|\le \left|f\left(x\right)\right|\left|g\left(x\right)-M\right|+\left|M\right|\left|f\left(x\right)-L\right|\text{.)}$

A function has to be continuous at $x=a$ if the $\underset{x\to a}{\text{lim}}f\left(x\right)$ exists.

You can use the quotient rule to evaluate $\underset{x\to 0}{\text{lim}}\frac{\text{sin}x}{x}.$

If there is a vertical asymptote at $x=a$ for the function $f\left(x\right),$ then f is undefined at the point $x=a.$

If $\underset{x\to a}{\text{lim}}f\left(x\right)$ does not exist, then f is undefined at the point $x=a.$

Using the graph, find each limit or explain why the limit does not exist.

1. $\underset{x\to \mathrm{-1}}{\text{lim}}f\left(x\right)$
2. $\underset{x\to 1}{\text{lim}}f\left(x\right)$
3. $\underset{x\to 0^{+}}{\text{lim}}f\left(x\right)$
4. $\underset{x\to 2}{\text{lim}}f\left(x\right)$

$\underset{x\to 2}{\text{lim}}\frac{2x^2-3x-2}{x-2}$

$\underset{x\to 0}{\text{lim}}3x^2-2x+4$

$\underset{x\to 3}{\text{lim}}\frac{x^3-2x^2-1}{3x-2}$

$\underset{x\to \pi \text{/}2}{\text{lim}}\frac{\text{cot}x}{\text{cos}x}$

$\underset{x\to \mathrm{-5}}{\text{lim}}\frac{x^2+25}{x+5}$

$\underset{x\to 2}{\text{lim}}\frac{3x^2-2x-8}{x^2-4}$

$\underset{x\to 1}{\text{lim}}\frac{x^2-1}{x^3-1}$

$\underset{x\to 1}{\text{lim}}\frac{x^2-1}{\sqrt{x}-1}$

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

$\underset{x\to 4}{\text{lim}}\frac{1}{\sqrt{x}-2}$

$\underset{x\to 0}{\text{lim}}{x}^2\text{cos}\left(2\pi x\right)=0$

$\underset{x\to 0}{\text{lim}}{x}^3\text{sin}\left(\frac{\pi }{x}\right)=0$

Determine the domain such that the function $f(x)=\sqrt{x-2}+x{e}^x$ is continuous over its domain.

$f\left(x\right)=\{\begin{array}{l}{x}^2+1,x>c\\ 2x,x\le c\end{array}$

$f\left(x\right)=\{\begin{array}{l}\sqrt{x+1},x>\text{−}1\\ x^2+c,x\le -1\end{array}$

$\underset{x\to 1}{\text{lim}}(8x+16)=24$

$\underset{x\to 0}{\text{lim}}{x}^3=0$

A ball is thrown into the air and the vertical position is given by $x\left(t\right)=\mathrm{-4.9}{t}^2+25t+5.$ Use the Intermediate Value Theorem to show that the ball must land on the ground sometime between 5 sec and 6 sec after the throw.

A particle moving along a line has a displacement according to the function $x\left(t\right)=t^2-2t+4,$ where x is measured in meters and t is measured in seconds. Find the average velocity over the time period $t=\left[0,2\right].$

From the previous exercises, estimate the instantaneous velocity at $t=2$ by checking the average velocity within $t=0.01\text{sec}\text{.}$