Mixing
if limf(x) = l and limg(x)=m prove that lim max{f,g}(x) = max{l,m}
$begingroup$
i have to prove the following :
if lim f(x) = l and lim g(x)=m
prove that
lim max{f,g}(x) = max{l,m}
any help about how to do it?
limits functions
asked Nov 12 '13 at 0:04
socratessocrates
112
$endgroup$
|
show 1 more comment
$begingroup$
i have to prove the following :
if lim f(x) = l and lim g(x)=m
prove that
lim max{f,g}(x) = max{l,m}
any help about how to do it?
limits functions
asked Nov 12 '13 at 0:04
socratessocrates
112
$endgroup$
$begingroup$
The limit as $x$ goes to what?
$endgroup$
– Wintermute
Nov 12 '13 at 0:29
$begingroup$
oh yes, it goes to a, x---> a
$endgroup$
– socrates
Nov 12 '13 at 0:32
$begingroup$
@mtiano i hope you can help,i'd be gratefull .....
$endgroup$
– socrates
Nov 12 '13 at 0:33
$begingroup$
This question does not make sense to me. If $f(x)=1$ and $g(x)=x$ then the limits as $x rightarrow 0$ are 1 and 0 respectively. However $max (f(x),g(x)) neq max (1,0)$
$endgroup$
– Wintermute
Nov 12 '13 at 0:35
$begingroup$
Sorry, had to edit my last. What are you taking your max over?
$endgroup$
– Wintermute
Nov 12 '13 at 0:37
|
show 1 more comment
$begingroup$
i have to prove the following :
if lim f(x) = l and lim g(x)=m
prove that
lim max{f,g}(x) = max{l,m}
any help about how to do it?
limits functions
asked Nov 12 '13 at 0:04
socratessocrates
112
$endgroup$
i have to prove the following :
if lim f(x) = l and lim g(x)=m
prove that
lim max{f,g}(x) = max{l,m}
any help about how to do it?
limits functions
limits functions
asked Nov 12 '13 at 0:04
socratessocrates
112
asked Nov 12 '13 at 0:04
socratessocrates
112
asked Nov 12 '13 at 0:04
socratessocrates
112
asked Nov 12 '13 at 0:04
socratessocrates
112
asked Nov 12 '13 at 0:04
socratessocrates
112
112
$begingroup$
The limit as $x$ goes to what?
$endgroup$
– Wintermute
Nov 12 '13 at 0:29
$begingroup$
oh yes, it goes to a, x---> a
$endgroup$
– socrates
Nov 12 '13 at 0:32
$begingroup$
@mtiano i hope you can help,i'd be gratefull .....
$endgroup$
– socrates
Nov 12 '13 at 0:33
$begingroup$
This question does not make sense to me. If $f(x)=1$ and $g(x)=x$ then the limits as $x rightarrow 0$ are 1 and 0 respectively. However $max (f(x),g(x)) neq max (1,0)$
$endgroup$
– Wintermute
Nov 12 '13 at 0:35
$begingroup$
Sorry, had to edit my last. What are you taking your max over?
$endgroup$
– Wintermute
Nov 12 '13 at 0:37
|
show 1 more comment
$begingroup$
The limit as $x$ goes to what?
$endgroup$
– Wintermute
Nov 12 '13 at 0:29
$begingroup$
oh yes, it goes to a, x---> a
$endgroup$
– socrates
Nov 12 '13 at 0:32
$begingroup$
@mtiano i hope you can help,i'd be gratefull .....
$endgroup$
– socrates
Nov 12 '13 at 0:33
$begingroup$
This question does not make sense to me. If $f(x)=1$ and $g(x)=x$ then the limits as $x rightarrow 0$ are 1 and 0 respectively. However $max (f(x),g(x)) neq max (1,0)$
$endgroup$
– Wintermute
Nov 12 '13 at 0:35
$begingroup$
Sorry, had to edit my last. What are you taking your max over?
$endgroup$
– Wintermute
Nov 12 '13 at 0:37
$begingroup$
The limit as $x$ goes to what?
$endgroup$
– Wintermute
Nov 12 '13 at 0:29
$begingroup$
The limit as $x$ goes to what?
$endgroup$
– Wintermute
Nov 12 '13 at 0:29
$begingroup$
oh yes, it goes to a, x---> a
$endgroup$
– socrates
Nov 12 '13 at 0:32
$begingroup$
oh yes, it goes to a, x---> a
$endgroup$
– socrates
Nov 12 '13 at 0:32
$begingroup$
@mtiano i hope you can help,i'd be gratefull .....
$endgroup$
– socrates
Nov 12 '13 at 0:33
$begingroup$
@mtiano i hope you can help,i'd be gratefull .....
$endgroup$
– socrates
Nov 12 '13 at 0:33
$begingroup$
This question does not make sense to me. If $f(x)=1$ and $g(x)=x$ then the limits as $x rightarrow 0$ are 1 and 0 respectively. However $max (f(x),g(x)) neq max (1,0)$
$endgroup$
– Wintermute
Nov 12 '13 at 0:35
$begingroup$
This question does not make sense to me. If $f(x)=1$ and $g(x)=x$ then the limits as $x rightarrow 0$ are 1 and 0 respectively. However $max (f(x),g(x)) neq max (1,0)$
$endgroup$
– Wintermute
Nov 12 '13 at 0:35
$begingroup$
Sorry, had to edit my last. What are you taking your max over?
$endgroup$
– Wintermute
Nov 12 '13 at 0:37
$begingroup$
Sorry, had to edit my last. What are you taking your max over?
$endgroup$
– Wintermute
Nov 12 '13 at 0:37
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
Hint: For any real $a,binBbb R,$ we have that $$max{a,b}=frac{a+b+|a-b|}2.$$ In particular, then, applying absolute value properties, we have $$begin{align}bigl|max{f(x),g(x)}-max{l,m}bigr| &= left|frac{bigl(f(x)-lbigr)+bigl(g(x)-mbigr)+|f(x)-g(x)|-|l-m|}2right|\ &le frac{|f(x)-l|}2+frac{|g(x)-m|}2+frac12bigl||f(x)-g(x)|-|l-m|bigr|.end{align}$$
Note also that $bigl||a|-|b|bigr|le |a-b|$ for all $a,binBbb R$ (readily proved from triangle inequality. See if you can take it from there.
answered Nov 12 '13 at 1:12
Cameron BuieCameron Buie
85.8k772160
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: For any real $a,binBbb R,$ we have that $$max{a,b}=frac{a+b+|a-b|}2.$$ In particular, then, applying absolute value properties, we have $$begin{align}bigl|max{f(x),g(x)}-max{l,m}bigr| &= left|frac{bigl(f(x)-lbigr)+bigl(g(x)-mbigr)+|f(x)-g(x)|-|l-m|}2right|\ &le frac{|f(x)-l|}2+frac{|g(x)-m|}2+frac12bigl||f(x)-g(x)|-|l-m|bigr|.end{align}$$
Note also that $bigl||a|-|b|bigr|le |a-b|$ for all $a,binBbb R$ (readily proved from triangle inequality. See if you can take it from there.
answered Nov 12 '13 at 1:12
Cameron BuieCameron Buie
85.8k772160
$endgroup$
add a comment |
$begingroup$
Hint: For any real $a,binBbb R,$ we have that $$max{a,b}=frac{a+b+|a-b|}2.$$ In particular, then, applying absolute value properties, we have $$begin{align}bigl|max{f(x),g(x)}-max{l,m}bigr| &= left|frac{bigl(f(x)-lbigr)+bigl(g(x)-mbigr)+|f(x)-g(x)|-|l-m|}2right|\ &le frac{|f(x)-l|}2+frac{|g(x)-m|}2+frac12bigl||f(x)-g(x)|-|l-m|bigr|.end{align}$$
Note also that $bigl||a|-|b|bigr|le |a-b|$ for all $a,binBbb R$ (readily proved from triangle inequality. See if you can take it from there.
answered Nov 12 '13 at 1:12
Cameron BuieCameron Buie
85.8k772160
$endgroup$
add a comment |
$begingroup$
Hint: For any real $a,binBbb R,$ we have that $$max{a,b}=frac{a+b+|a-b|}2.$$ In particular, then, applying absolute value properties, we have $$begin{align}bigl|max{f(x),g(x)}-max{l,m}bigr| &= left|frac{bigl(f(x)-lbigr)+bigl(g(x)-mbigr)+|f(x)-g(x)|-|l-m|}2right|\ &le frac{|f(x)-l|}2+frac{|g(x)-m|}2+frac12bigl||f(x)-g(x)|-|l-m|bigr|.end{align}$$
Note also that $bigl||a|-|b|bigr|le |a-b|$ for all $a,binBbb R$ (readily proved from triangle inequality. See if you can take it from there.
answered Nov 12 '13 at 1:12
Cameron BuieCameron Buie
85.8k772160
$endgroup$
Hint: For any real $a,binBbb R,$ we have that $$max{a,b}=frac{a+b+|a-b|}2.$$ In particular, then, applying absolute value properties, we have $$begin{align}bigl|max{f(x),g(x)}-max{l,m}bigr| &= left|frac{bigl(f(x)-lbigr)+bigl(g(x)-mbigr)+|f(x)-g(x)|-|l-m|}2right|\ &le frac{|f(x)-l|}2+frac{|g(x)-m|}2+frac12bigl||f(x)-g(x)|-|l-m|bigr|.end{align}$$
Note also that $bigl||a|-|b|bigr|le |a-b|$ for all $a,binBbb R$ (readily proved from triangle inequality. See if you can take it from there.
answered Nov 12 '13 at 1:12
Cameron BuieCameron Buie
85.8k772160
answered Nov 12 '13 at 1:12
Cameron BuieCameron Buie
85.8k772160
answered Nov 12 '13 at 1:12
Cameron BuieCameron Buie
85.8k772160
answered Nov 12 '13 at 1:12
Cameron BuieCameron Buie
85.8k772160
85.8k772160
add a comment |
add a comment |
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$begingroup$
The limit as $x$ goes to what?
$endgroup$
– Wintermute
Nov 12 '13 at 0:29
$begingroup$
oh yes, it goes to a, x---> a
$endgroup$
– socrates
Nov 12 '13 at 0:32
$begingroup$
@mtiano i hope you can help,i'd be gratefull .....
$endgroup$
– socrates
Nov 12 '13 at 0:33
$begingroup$
This question does not make sense to me. If $f(x)=1$ and $g(x)=x$ then the limits as $x rightarrow 0$ are 1 and 0 respectively. However $max (f(x),g(x)) neq max (1,0)$
$endgroup$
– Wintermute
Nov 12 '13 at 0:35
$begingroup$
Sorry, had to edit my last. What are you taking your max over?
$endgroup$
– Wintermute
Nov 12 '13 at 0:37