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Use a table of numbers to investigate lim.
The table suggests \mathop {\lim\limits_{x \to 0}} \dfrac{1 - \cos x}{1 + \cos x} = 0.
\xrightarrow{x\ {\rm approaches}\ 0\atop \mbox{from the left}} | \xleftarrow{x\ {\rm approaches}\ 0\atop \mbox{from the right}} | ||||||
---|---|---|---|---|---|---|---|
x | -0.1 | -0.01 | -0.001 | \rightarrow 0 \leftarrow | 0.001 | 0.01 | 0.1 |
f(x)=\dfrac{1 - \cos x}{1 + \cos x} | 0.002504 | 0.000025 | 0.00000025 | f(x) approaches 0 | 0.00000025 | 0.000025 | 0.002504 |
In Problems 2 and 3, use a graph to investigate \lim\limits_{x\rightarrow c}f(x).
f(x)=\left\{ \begin{array}{l@{\quad}l@{\quad}l} 2x-5 & & \hbox{if }x\lt1 \\ 6-9x & & \hbox{if }x\geq 1 \end{array} \right. \quad at c=1
f(x)=\left\{ \begin{array}{l@{\quad}l@{\quad}l} x^{2}+2 & & \hbox{if }x\lt2 \\ 2x+1 & & \hbox{if }x\geq 2 \end{array} \right. \quad at c=2
\mathop {\lim\limits_{x \to 2}} f(x) does not exist.
For f(x)=x^{2}-3:
In Problems 5 and 6, for each function find the limit of the difference quotient \lim\limits_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}.
f(x)=\dfrac{3}{x}
- \dfrac{3}{x^2}
f(x)=3x^{2}+2x
Find \lim\limits_{x\rightarrow 0}f(x) if 1+\sin x\leq f(x)\leq \vert x\vert +1
1
In Problems 8-22, find each limit.
\lim\limits_{x\rightarrow 2} \left( 2x-\dfrac{1}{x}\right)
\lim\limits_{x\rightarrow \pi } \left( x\cos x\right)
-\pi
\lim\limits_{x\rightarrow -1}\left( x^{3}+3x^{2}-x-1\right)
\lim\limits_{x\rightarrow 0}\sqrt[3]{x( x+2) ^{3}}
0
{\lim\limits_{x\rightarrow 0}} [(2x+3) (x^{5}+5x)]
{\lim\limits_{x\rightarrow 3}} \dfrac{x^{3}-27}{x-3}
27
{\lim\limits_{x\rightarrow 3}}\left(\dfrac{x^{2}}{x-3}-\dfrac{3x}{x-3}\right)
{\lim\limits_{x\rightarrow 2}}\dfrac{x^{2}-4}{x-2}
4
{\lim\limits_{x\rightarrow -1}}\dfrac{x^{2}+3x+2} {x^{2}+4x+3}
{\lim\limits_{x\rightarrow -2}} \dfrac{x^{3}+5x^{2}+6x}{x^{2}+x-2}
\dfrac{2}{3}
{\lim\limits_{x\rightarrow 1}} \left(x^{2}-3x+\dfrac{1}{x}\right) ^{15}
{\lim\limits_{x\rightarrow 2}}\dfrac{3-\sqrt{x^{2}+5}}{x^{2}-4}
- \dfrac{1}{6}
{\lim\limits_{x\rightarrow 0}} \left\{ \dfrac{1}{x}\left[ \dfrac{1}{(2+x)^{2}}-\dfrac{1}{4}\right] \right\}
{\lim\limits_{x\rightarrow 0}} \dfrac{\left({x+3}\right) ^{2}-9}{x}
6
{\lim\limits_{x\rightarrow 1}} [(x^{3}-3x^{2}+3x-1) (x+1) ^{2}]
In Problems 23-28, find each one-sided limit, if it exists.
{\lim\limits_{x\rightarrow - 2^{+}}}\dfrac{ x^{2}+5x+6}{x+2}
5
{\lim\limits_{x \rightarrow 5^{+}}}\dfrac{|x-5|}{x-5}
{\lim\limits_{x\rightarrow 1^{-}}}\dfrac{|x-1|}{x-1}
-1
{\lim\limits_{x\rightarrow \,3/2^{+}}}\lfloor 2x\rfloor
{\lim\limits_{x\rightarrow 4^{-}}}\dfrac{x^{2}-16}{x-4}
8
{\lim\limits_{x\rightarrow 1^{+}}},\sqrt{x-1}
In Problems 29 and 30, find \lim\limits_{x\rightarrow c^{-}}f(x) and \lim\limits_{x\rightarrow c^{+}}f(x) for the given c Determine whether \lim\limits_{x\rightarrow c}f(x) exists.
f(x)=\left\{ \begin{array}{l@{\quad}l} 2x+3 & \hbox{if }x\lt 2 \\[3pt] 9-x & \hbox{if }x\geq 2 \end{array} \right. at c=2
\mathop{\lim\limits_{x \to 2^ - }} f(x) = 7,\mathop{\lim\limits_{x \to 2^ + }} f(x) = 7, \mathop{\lim\limits_{x \to 2}} f(x) = 7
142
f(x)=\left\{ \begin{array}{c@{ }c} 3x+1 & \hbox{if }x \lt 3 \\[3pt] 10 & \hbox{if }x=3 \\[3pt] 4x-2 & \hbox{if }x \gt 3 \end{array} \right. at c=3
In Problems 31-36, determine whether f is continuous at c.
f(x)=\left\{ \begin{array}{c@{}c} 5x-2 & \hbox{if }x \lt 1 \\[3pt] 5 & \hbox{if }x=1 \\[3pt] 2x+1 & \hbox{if }x \gt 1 \end{array} \right. at c=1
f is not continuous at c = 1.
f(x)=\left\{ \begin{array}{c@{}c} x^{2} & \hbox{if }x \lt -1 \\[3pt] 2 & \hbox{if }x=-1 \\[3pt] -3x-2 & \hbox{if }x \gt -1 \end{array} \right. at c=-1
f(x)=\left\{ \begin{array}{c@{}l} 4-3x^{2} & \hbox{if }x\lt0 \\[3pt] 4 & \hbox{if }x=0 \\[3pt] \sqrt{16-x^{2}} & \hbox{if }0 \lt x\leq 4 \end{array} \right. at c=0
f is continuous at c = 0.
f(x)=\left\{ \begin{array}{c@{}l} \sqrt{4+x} & \hbox{if }-4\leq x\leq 4 \\[3pt] \sqrt{\dfrac{x^{2}-16}{x-4}} & \hbox{if }x>4 \end{array} \right. at c=4
f(x) = \lfloor \,2x\rfloor at c=\dfrac{1}{ 2}
f is not continuous at c=\raise.5ex\hbox{\(\scriptstyle 1\)}\kern-.1em/ \kern-.15em\lower.25ex\hbox{\(\scriptstyle 2\)}.
f(x)=|\,x-5\,| at c=5
A function f is defined on the interval [-1,1] with the following properties: f is continuous on [-1,1] except at 0, negative at -1, positive at 1, but with no zeros. Does this contradict the Intermediate Value theorem?
In Problems 39-43 find all values x for which f(x) is continuous.
f(x)=\dfrac{x}{x^{3}-27}
f is continuous on the set \{ x \vert x \ne 3 {\}}.
f(x)=\dfrac{x^{2}-3}{x^{2}+5x+6}
f(x)=\dfrac{2x+1}{x^{3}+4x^{2}+4x}
f is continuous on the set \{ x \vert x \ne -2, x \ne 0 \}.
f(x) =\sqrt{x-1}
f(x) =2^{-x}
f is continuous on the set of all real numbers.
Use the Intermediate Value Theorem to determine whether 2x^{3}+3x^{2}-23x-42=0 has a zero in the interval [3,4].
In Problems 45 and 46, use the Intermediate Value Theorem to approximate the zero correct to three decimal places.
f(x) =8x^{4}-2x^{2}+5x-1 on the interval \left[0,1\right].
0.215
f(x) =3x^{3}-10x+9; zero between -3 and -2.
Find {\lim\limits_{x\rightarrow 0^{+}}}\dfrac{|\,x\,|}{x}(1-x) and {\lim\limits_{x\rightarrow 0^{-}}}\dfrac{|\,x\,|}{x}(1-x). What can you say about {\lim\limits_{x\rightarrow 0}}\dfrac{|x|}{x} (1-x)?
\mathop {\lim\limits_{x \to 0^ + }} \dfrac{| x |}{x}(1 - x) = 1, \mathop{\lim \limits_{x \to 0^ - }} \dfrac{| x |}{x}(1 - x) = - 1, \mathop{\lim \limits_{x \to 0}} \dfrac{| x |}{x}(1 - x) does not exist.
Find {\lim\limits_{x\rightarrow 2}}\left( \dfrac{x^{2}}{x-2}- \dfrac{2x}{x-2}\right). Then comment on the statement that this limit is given by \lim\limits_{x\rightarrow 2}\dfrac{x^{2}}{x-2}-\lim\limits_{x\rightarrow 2}\dfrac{2x}{x-2}.
Find {\lim\limits_{h\rightarrow 0}} \dfrac{f(x+h)-f(x)}{h} for f(x)=\sqrt{x}.
\dfrac{1}{2\sqrt x}
For \lim\limits_{x\rightarrow 3}(2x+1)=7, find the largest possible \delta that “works” for \epsilon =0.01.
In Problems 51-60, find each limit.
\lim\limits_{x\rightarrow 0}\cos (tan x)
1
\lim\limits_{x\rightarrow 0}{\dfrac{{\sin {\dfrac{{x}}{{4}}}}}{{x}}}
\lim\limits_{x\rightarrow 0}\,\dfrac{\tan (3x) }{\tan ( 4x) }
\dfrac{3}{4}
{\lim\limits_{x\rightarrow 0}}\dfrac{\cos {\dfrac{x}{3}-1}}{x}
\lim\limits_{x\rightarrow 0}\left( \dfrac{\cos x-1}{x}\right) ^{10}
0
\lim\limits_{x\rightarrow 0}{\dfrac{{e^{4x}-1}}{e^{x}{-1}}}
\lim\limits_{x\rightarrow \pi /2^{+}}\tan x
-\infty
\lim\limits_{x\rightarrow -3}\dfrac{2+x}{( x+3) ^{2}}
\lim\limits_{x\rightarrow \infty }\dfrac{3x^{3}-2x+1}{x^{3}-8}
3
\lim\limits_{x\rightarrow \infty }\dfrac{3x^{4}+x}{2x^{2}}
In Problems 61 and 62, find any vertical and horizontal asymptotes of f.
f(x)=\dfrac{4x-2}{x+3}
x = -3 is a vertical asymptote. y = 4 is a horizontal asymptote.
f(x)=\dfrac{2x}{x^{2}-4}
Let f( x) =\left\{ \begin{array}{c@{}c} \dfrac{\tan x}{2x} & \hbox{if }x\neq 0 \\ \dfrac{1}{2} & \hbox{if }x=0 \end{array} \right.. Is f continuous at 0?
Yes
Let f( x) =\left\{ \begin{array}{c@{}c} \dfrac{\sin ( 3x) }{x} & \hbox{if }x\neq 0 \\ 1 & \hbox{if }x=0 \end{array} \right.. Is f continuous at 0?
The function f( x) =\dfrac{\cos \left( \pi x+\dfrac{\pi }{2}\right) }{x} is not defined at 0. Decide how to define f( 0) so that f is continuous at 0.
f(0) = -\pi
Use an \epsilon - \delta argument to show that the statement \lim\limits_{x\rightarrow -3} (x^{2}-9) =-18 is false.
If 1-x^{2}\leq f(x)\leq cos x for all x in the interval -\dfrac{\pi }{2} \lt x \lt \dfrac{\pi }{2}, show that \lim\limits_{x\rightarrow 0}f(x)=1.
See Student Solution Manual.