$lim_{xto 0}frac{f(ax)}{x},$ knowing the limit of $frac{f(x)}{x}$
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 yesterday
amWhy
192k28224439
asked yesterday
Jules
1067
add a comment |
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 yesterday
amWhy
192k28224439
asked yesterday
Jules
1067
3
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
– user3482749
yesterday
add a comment |
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 yesterday
amWhy
192k28224439
asked yesterday
Jules
1067
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 yesterday
amWhy
192k28224439
asked yesterday
Jules
1067
edited yesterday
amWhy
192k28224439
asked yesterday
Jules
1067
edited yesterday
amWhy
192k28224439
edited yesterday
amWhy
192k28224439
edited yesterday
amWhy
192k28224439
192k28224439
asked yesterday
Jules
1067
asked yesterday
Jules
1067
asked yesterday
Jules
1067
1067
3
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
– user3482749
yesterday
add a comment |
3
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
– user3482749
yesterday
3
3
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
– user3482749
yesterday
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
– user3482749
yesterday
add a comment |
4 Answers
4
active
oldest
votes
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered yesterday
Henry
98.2k475161
add a comment |
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited 21 hours ago
Math_QED
7,11731449
answered yesterday
Tsemo Aristide
56.1k11444
add a comment |
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 yesterday
Mostafa Ayaz
13.7k3936
add a comment |
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 yesterday
Math_QED
7,11731449
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
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered yesterday
Henry
98.2k475161
add a comment |
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered yesterday
Henry
98.2k475161
add a comment |
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered yesterday
Henry
98.2k475161
Hint:
Consider $$limlimits_{x rightarrow 0}{frac{f(ax)}{ax}}$$
answered yesterday
Henry
98.2k475161
answered yesterday
Henry
98.2k475161
answered yesterday
Henry
98.2k475161
answered yesterday
Henry
98.2k475161
98.2k475161
add a comment |
add a comment |
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited 21 hours ago
Math_QED
7,11731449
answered yesterday
Tsemo Aristide
56.1k11444
add a comment |
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited 21 hours ago
Math_QED
7,11731449
answered yesterday
Tsemo Aristide
56.1k11444
add a comment |
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited 21 hours ago
Math_QED
7,11731449
answered yesterday
Tsemo Aristide
56.1k11444
$$lim_{xrightarrow 0}{{f(ax)}over x}=lim_{xrightarrow 0}a{{f(ax)}over {ax}}=ac$$
edited 21 hours ago
Math_QED
7,11731449
answered yesterday
Tsemo Aristide
56.1k11444
edited 21 hours ago
Math_QED
7,11731449
edited 21 hours ago
Math_QED
7,11731449
edited 21 hours ago
Math_QED
7,11731449
7,11731449
answered yesterday
Tsemo Aristide
56.1k11444
answered yesterday
Tsemo Aristide
56.1k11444
answered yesterday
Tsemo Aristide
56.1k11444
56.1k11444
add a comment |
add a comment |
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 yesterday
Mostafa Ayaz
13.7k3936
add a comment |
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 yesterday
Mostafa Ayaz
13.7k3936
add a comment |
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 yesterday
Mostafa Ayaz
13.7k3936
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 yesterday
Mostafa Ayaz
13.7k3936
answered yesterday
Mostafa Ayaz
13.7k3936
answered yesterday
Mostafa Ayaz
13.7k3936
answered yesterday
Mostafa Ayaz
13.7k3936
13.7k3936
add a comment |
add a comment |
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 yesterday
Math_QED
7,11731449
add a comment |
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 yesterday
Math_QED
7,11731449
add a comment |
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 yesterday
Math_QED
7,11731449
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 yesterday
Math_QED
7,11731449
answered yesterday
Math_QED
7,11731449
answered yesterday
Math_QED
7,11731449
answered yesterday
Math_QED
7,11731449
7,11731449
add a comment |
add a comment |
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3
Hint: defining $y = ax$, we have $limlimits_{xto 0}frac{f(ax)}{x} = limlimits_{yto 0}frac{f(y)}{y/a}$.
– user3482749
yesterday