Mixing
$lim_{xto 0}frac{f(ax)}{x},$ knowing the limit of $frac{f(x)}{x}$
$begingroup$
If we know that
$$lim_{x rightarrow 0}{frac{f(x)}{x}}rightarrow c,; text{ for a given }cinmathbb{R}.$$ Show that
$$
lim_{x rightarrow 0}{frac{f(ax)}{x}}rightarrow ac; text{ for any } ainmathbb{R}setminus{{0}}.
$$
We definitely know from the above (using the product rule for limits)
$$
lim_{x rightarrow 0}{f(x)}rightarrow 0 ;;text{ holds. }
$$
Don't know if that helps.
calculus limits
edited Jan 2 at 20:15
amWhy
192k28225439
asked Jan 2 at 19:55
JulesJules
1087
$endgroup$
add a comment |
$begingroup$
If we know that
$$lim_{x rightarrow 0}{frac{f(x)}{x}}rightarrow c,; text{ for a given }cinmathbb{R}.$$ Show that
$$
lim_{x rightarrow 0}{frac{f(ax)}{x}}rightarrow ac; text{ for any } ainmathbb{R}setminus{{0}}.
$$
We definitely know from the above (using the product rule for limits)
$$
lim_{x rightarrow 0}{f(x)}rightarrow 0 ;;text{ holds. }
$$
Don't know if that helps.
calculus limits
edited Jan 2 at 20:15
amWhy
192k28225439
asked Jan 2 at 19:55
JulesJules
1087
$endgroup$
3
$begingroup$
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
$endgroup$
– user3482749
Jan 2 at 19:58
add a comment |
$begingroup$
If we know that
$$lim_{x rightarrow 0}{frac{f(x)}{x}}rightarrow c,; text{ for a given }cinmathbb{R}.$$ Show that
$$
lim_{x rightarrow 0}{frac{f(ax)}{x}}rightarrow ac; text{ for any } ainmathbb{R}setminus{{0}}.
$$
We definitely know from the above (using the product rule for limits)
$$
lim_{x rightarrow 0}{f(x)}rightarrow 0 ;;text{ holds. }
$$
Don't know if that helps.
calculus limits
edited Jan 2 at 20:15
amWhy
192k28225439
asked Jan 2 at 19:55
JulesJules
1087
$endgroup$
If we know that
$$lim_{x rightarrow 0}{frac{f(x)}{x}}rightarrow c,; text{ for a given }cinmathbb{R}.$$ Show that
$$
lim_{x rightarrow 0}{frac{f(ax)}{x}}rightarrow ac; text{ for any } ainmathbb{R}setminus{{0}}.
$$
We definitely know from the above (using the product rule for limits)
$$
lim_{x rightarrow 0}{f(x)}rightarrow 0 ;;text{ holds. }
$$
Don't know if that helps.
calculus limits
calculus limits
edited Jan 2 at 20:15
amWhy
192k28225439
asked Jan 2 at 19:55
JulesJules
1087
edited Jan 2 at 20:15
amWhy
192k28225439
asked Jan 2 at 19:55
JulesJules
1087
edited Jan 2 at 20:15
amWhy
192k28225439
edited Jan 2 at 20:15
amWhy
192k28225439
edited Jan 2 at 20:15
amWhy
192k28225439
192k28225439
asked Jan 2 at 19:55
JulesJules
1087
asked Jan 2 at 19:55
JulesJules
1087
asked Jan 2 at 19:55
JulesJules
1087
1087
3
$begingroup$
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
$endgroup$
– user3482749
Jan 2 at 19:58
add a comment |
3
$begingroup$
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
$endgroup$
– user3482749
Jan 2 at 19:58
3
3
$begingroup$
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
$endgroup$
– user3482749
Jan 2 at 19:58
$begingroup$
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
$endgroup$
– user3482749
Jan 2 at 19:58
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
As this is elementary, this may require an $epsilon-delta$ argument.
We want to make the following quantity small:
$$left|frac{f(ax)}{x}-acright|$$
Observe that
$$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$
and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.
answered Jan 2 at 20:17
Math_QEDMath_QED
7,33631450
$endgroup$
add a comment |
$begingroup$
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered Jan 2 at 19:58
HenryHenry
98.7k476163
$endgroup$
add a comment |
$begingroup$
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited Jan 3 at 8:28
Math_QED
7,33631450
answered Jan 2 at 19:58
Tsemo AristideTsemo Aristide
56.8k11444
$endgroup$
add a comment |
$begingroup$
Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$
answered Jan 2 at 20:02
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As this is elementary, this may require an $epsilon-delta$ argument.
We want to make the following quantity small:
$$left|frac{f(ax)}{x}-acright|$$
Observe that
$$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$
and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.
answered Jan 2 at 20:17
Math_QEDMath_QED
7,33631450
$endgroup$
add a comment |
$begingroup$
As this is elementary, this may require an $epsilon-delta$ argument.
We want to make the following quantity small:
$$left|frac{f(ax)}{x}-acright|$$
Observe that
$$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$
and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.
answered Jan 2 at 20:17
Math_QEDMath_QED
7,33631450
$endgroup$
add a comment |
$begingroup$
As this is elementary, this may require an $epsilon-delta$ argument.
We want to make the following quantity small:
$$left|frac{f(ax)}{x}-acright|$$
Observe that
$$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$
and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.
answered Jan 2 at 20:17
Math_QEDMath_QED
7,33631450
$endgroup$
As this is elementary, this may require an $epsilon-delta$ argument.
We want to make the following quantity small:
$$left|frac{f(ax)}{x}-acright|$$
Observe that
$$left|frac{f(ax)}{x}-acright|= left|frac{f(ax)}{ax}a-acright|= |a|left|frac{f(ax)}{ax}-cright|$$
and this last factor can be made small because $f(x)/x to c$. It is a good exercise on this level to make this into a rigorous proof.
answered Jan 2 at 20:17
Math_QEDMath_QED
7,33631450
answered Jan 2 at 20:17
Math_QEDMath_QED
7,33631450
answered Jan 2 at 20:17
Math_QEDMath_QED
7,33631450
answered Jan 2 at 20:17
Math_QEDMath_QED
7,33631450
7,33631450
add a comment |
add a comment |
$begingroup$
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered Jan 2 at 19:58
HenryHenry
98.7k476163
$endgroup$
add a comment |
$begingroup$
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered Jan 2 at 19:58
HenryHenry
98.7k476163
$endgroup$
add a comment |
$begingroup$
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered Jan 2 at 19:58
HenryHenry
98.7k476163
$endgroup$
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered Jan 2 at 19:58
HenryHenry
98.7k476163
answered Jan 2 at 19:58
HenryHenry
98.7k476163
answered Jan 2 at 19:58
HenryHenry
98.7k476163
answered Jan 2 at 19:58
HenryHenry
98.7k476163
98.7k476163
add a comment |
add a comment |
$begingroup$
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited Jan 3 at 8:28
Math_QED
7,33631450
answered Jan 2 at 19:58
Tsemo AristideTsemo Aristide
56.8k11444
$endgroup$
add a comment |
$begingroup$
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited Jan 3 at 8:28
Math_QED
7,33631450
answered Jan 2 at 19:58
Tsemo AristideTsemo Aristide
56.8k11444
$endgroup$
add a comment |
$begingroup$
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited Jan 3 at 8:28
Math_QED
7,33631450
answered Jan 2 at 19:58
Tsemo AristideTsemo Aristide
56.8k11444
$endgroup$
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited Jan 3 at 8:28
Math_QED
7,33631450
answered Jan 2 at 19:58
Tsemo AristideTsemo Aristide
56.8k11444
edited Jan 3 at 8:28
Math_QED
7,33631450
edited Jan 3 at 8:28
Math_QED
7,33631450
edited Jan 3 at 8:28
Math_QED
7,33631450
7,33631450
answered Jan 2 at 19:58
Tsemo AristideTsemo Aristide
56.8k11444
answered Jan 2 at 19:58
Tsemo AristideTsemo Aristide
56.8k11444
answered Jan 2 at 19:58
Tsemo AristideTsemo Aristide
56.8k11444
56.8k11444
add a comment |
add a comment |
$begingroup$
Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$
answered Jan 2 at 20:02
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
add a comment |
$begingroup$
Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$
answered Jan 2 at 20:02
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
add a comment |
$begingroup$
Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$
answered Jan 2 at 20:02
Mostafa AyazMostafa Ayaz
15.2k3939
$endgroup$
Let $$lim_{xto 0}{f(x)over x}=l$$therefore $$lim_{xto 0}{f(ax)over x}=lim_{xto 0}a{f(ax)over ax}=alim_{xto 0}{f(x)over x}=al$$
answered Jan 2 at 20:02
Mostafa AyazMostafa Ayaz
15.2k3939
answered Jan 2 at 20:02
Mostafa AyazMostafa Ayaz
15.2k3939
answered Jan 2 at 20:02
Mostafa AyazMostafa Ayaz
15.2k3939
answered Jan 2 at 20:02
Mostafa AyazMostafa Ayaz
15.2k3939
15.2k3939
add a comment |
add a comment |
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$begingroup$
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
$endgroup$
– user3482749
Jan 2 at 19:58