Double limit $lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 f(x) - x^2 f(y)}{(1-cos x)sin (x-y)}$Evaluate $...
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Double limit $lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 f(x) - x^2 f(y)}{(1-cos x)sin (x-y)}$
Evaluate $ lim_{xrightarrow{fracpi2 }} (sec(x) tan(x))^{cos(x)}$ without L'Hôpital's ruleLimit with arctan: $lim_{xrightarrow 0} frac{xsin 3x}{arctan x^2}$$lim_{xrightarrow0}frac{x^{672}-2cosleft(x^{1007}right)-sin(x^{672})+2}{x^{2014}}$Analyze if the limit exists: $lim_{xrightarrow0, x>0}sqrt{x}cdot sinleft(frac{1}{x}right)$Is it enough to show that $lim_{xrightarrow 0}cos(1/x)$ doesn't exist to show that $lim_{x rightarrow0}(2xsin(1/x)-cos(1/x))$ doesn't exist?Find $lim_{xrightarrow frac{pi}{4}}left(frac{sin x}{cos x}right)^left(frac{sin 2x}{cos 2x}right)$Does $lim_{xrightarrow 0}frac{ln(x)}{cot(x)}$ exist or not?Evaluating $lim_{xrightarrow 0}frac{cos(ax) - cos(bx)}{sin^2(x)}$Find $lim_{xrightarrowfrac{pi}{6}}frac{1-2sin x}{cos3x}$Computing the limit of $lim_{trightarrow0}tf(g(t))$ assuming $g(0)=0$ and $g'(0)>0$
$begingroup$
Let $I$ be an open interval containing $0$, and $f: I rightarrow R$ be differentiable and $f'$ be continuous. Compute:
$$lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 f(x) - x^2 f(y)}{(1-cos x)sin (x-y)}$$
This is my work:
$$lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 x^2 (frac{f(x)}{x^2}-frac{f(y)}{y^2})(x-y)}{(1-cos x)(x-y)sin(x-y)}$$
$$ lim_{xrightarrow0} frac{x^4}{1 - cos x} frac{xf'(x) - 2f(x)}{x^3}$$
$$ lim_{xrightarrow0}frac{x}{1-cos x} (xf'(x) - 2f(x)) $$
But it doesn't feel right because I'm "missing" one power of x to have $frac{x}{1-cos x}$ converge. Can anyone help me?
real-analysis calculus limits
asked yesterday
Bag of ChipsBag of Chips
1187
$endgroup$
add a comment |
$begingroup$
Let $I$ be an open interval containing $0$, and $f: I rightarrow R$ be differentiable and $f'$ be continuous. Compute:
$$lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 f(x) - x^2 f(y)}{(1-cos x)sin (x-y)}$$
This is my work:
$$lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 x^2 (frac{f(x)}{x^2}-frac{f(y)}{y^2})(x-y)}{(1-cos x)(x-y)sin(x-y)}$$
$$ lim_{xrightarrow0} frac{x^4}{1 - cos x} frac{xf'(x) - 2f(x)}{x^3}$$
$$ lim_{xrightarrow0}frac{x}{1-cos x} (xf'(x) - 2f(x)) $$
But it doesn't feel right because I'm "missing" one power of x to have $frac{x}{1-cos x}$ converge. Can anyone help me?
real-analysis calculus limits
asked yesterday
Bag of ChipsBag of Chips
1187
$endgroup$
$begingroup$
It seems that some condition like $f(0)=0$ is actually missing instead.
$endgroup$
– Saad
yesterday
$begingroup$
@Saad I forgot to add: derivative of f is continuous.
$endgroup$
– Bag of Chips
yesterday
$begingroup$
Please explain how you evaluate the inner limit. I get the same result but without the term involving $f(x)$.
$endgroup$
– maxmilgram
yesterday
1
$begingroup$
The power of $x$ in $dfrac x{1-cos x}simdfrac2x$ is indeed correct, but $limlimits_{x→0}dfrac{f(x)}x$ does not necessarily exist.
$endgroup$
– Saad
yesterday
$begingroup$
@maxmilgram I make a difference quotient by force to make use of assumption that f is differentiable. $lim_{y rightarrow x} frac{h(x) - h(y)}{x-y}$ = $h'(x)$ where $h(x) = frac{f(x)}{x^2}$ Also I make use of standard limit $lim_{z rightarrow 0} frac{sin z}{z} = 1$. But
$endgroup$
– Bag of Chips
yesterday
add a comment |
$begingroup$
Let $I$ be an open interval containing $0$, and $f: I rightarrow R$ be differentiable and $f'$ be continuous. Compute:
$$lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 f(x) - x^2 f(y)}{(1-cos x)sin (x-y)}$$
This is my work:
$$lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 x^2 (frac{f(x)}{x^2}-frac{f(y)}{y^2})(x-y)}{(1-cos x)(x-y)sin(x-y)}$$
$$ lim_{xrightarrow0} frac{x^4}{1 - cos x} frac{xf'(x) - 2f(x)}{x^3}$$
$$ lim_{xrightarrow0}frac{x}{1-cos x} (xf'(x) - 2f(x)) $$
But it doesn't feel right because I'm "missing" one power of x to have $frac{x}{1-cos x}$ converge. Can anyone help me?
real-analysis calculus limits
asked yesterday
Bag of ChipsBag of Chips
1187
$endgroup$
Let $I$ be an open interval containing $0$, and $f: I rightarrow R$ be differentiable and $f'$ be continuous. Compute:
$$lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 f(x) - x^2 f(y)}{(1-cos x)sin (x-y)}$$
This is my work:
$$lim_{xrightarrow0} lim_{yrightarrow x} frac{y^2 x^2 (frac{f(x)}{x^2}-frac{f(y)}{y^2})(x-y)}{(1-cos x)(x-y)sin(x-y)}$$
$$ lim_{xrightarrow0} frac{x^4}{1 - cos x} frac{xf'(x) - 2f(x)}{x^3}$$
$$ lim_{xrightarrow0}frac{x}{1-cos x} (xf'(x) - 2f(x)) $$
But it doesn't feel right because I'm "missing" one power of x to have $frac{x}{1-cos x}$ converge. Can anyone help me?
real-analysis calculus limits
real-analysis calculus limits
asked yesterday
Bag of ChipsBag of Chips
1187
asked yesterday
Bag of ChipsBag of Chips
1187
edited yesterday
Bag of Chips
asked yesterday
Bag of ChipsBag of Chips
1187
asked yesterday
Bag of ChipsBag of Chips
1187
asked yesterday
Bag of ChipsBag of Chips
1187
1187
$begingroup$
It seems that some condition like $f(0)=0$ is actually missing instead.
$endgroup$
– Saad
yesterday
$begingroup$
@Saad I forgot to add: derivative of f is continuous.
$endgroup$
– Bag of Chips
yesterday
$begingroup$
Please explain how you evaluate the inner limit. I get the same result but without the term involving $f(x)$.
$endgroup$
– maxmilgram
yesterday
1
$begingroup$
The power of $x$ in $dfrac x{1-cos x}simdfrac2x$ is indeed correct, but $limlimits_{x→0}dfrac{f(x)}x$ does not necessarily exist.
$endgroup$
– Saad
yesterday
$begingroup$
@maxmilgram I make a difference quotient by force to make use of assumption that f is differentiable. $lim_{y rightarrow x} frac{h(x) - h(y)}{x-y}$ = $h'(x)$ where $h(x) = frac{f(x)}{x^2}$ Also I make use of standard limit $lim_{z rightarrow 0} frac{sin z}{z} = 1$. But
$endgroup$
– Bag of Chips
yesterday
add a comment |
$begingroup$
It seems that some condition like $f(0)=0$ is actually missing instead.
$endgroup$
– Saad
yesterday
$begingroup$
@Saad I forgot to add: derivative of f is continuous.
$endgroup$
– Bag of Chips
yesterday
$begingroup$
Please explain how you evaluate the inner limit. I get the same result but without the term involving $f(x)$.
$endgroup$
– maxmilgram
yesterday
1
$begingroup$
The power of $x$ in $dfrac x{1-cos x}simdfrac2x$ is indeed correct, but $limlimits_{x→0}dfrac{f(x)}x$ does not necessarily exist.
$endgroup$
– Saad
yesterday
$begingroup$
@maxmilgram I make a difference quotient by force to make use of assumption that f is differentiable. $lim_{y rightarrow x} frac{h(x) - h(y)}{x-y}$ = $h'(x)$ where $h(x) = frac{f(x)}{x^2}$ Also I make use of standard limit $lim_{z rightarrow 0} frac{sin z}{z} = 1$. But
$endgroup$
– Bag of Chips
yesterday
$begingroup$
It seems that some condition like $f(0)=0$ is actually missing instead.
$endgroup$
– Saad
yesterday
$begingroup$
It seems that some condition like $f(0)=0$ is actually missing instead.
$endgroup$
– Saad
yesterday
$begingroup$
@Saad I forgot to add: derivative of f is continuous.
$endgroup$
– Bag of Chips
yesterday
$begingroup$
@Saad I forgot to add: derivative of f is continuous.
$endgroup$
– Bag of Chips
yesterday
$begingroup$
Please explain how you evaluate the inner limit. I get the same result but without the term involving $f(x)$.
$endgroup$
– maxmilgram
yesterday
$begingroup$
Please explain how you evaluate the inner limit. I get the same result but without the term involving $f(x)$.
$endgroup$
– maxmilgram
yesterday
1
1
$begingroup$
The power of $x$ in $dfrac x{1-cos x}simdfrac2x$ is indeed correct, but $limlimits_{x→0}dfrac{f(x)}x$ does not necessarily exist.
$endgroup$
– Saad
yesterday
$begingroup$
The power of $x$ in $dfrac x{1-cos x}simdfrac2x$ is indeed correct, but $limlimits_{x→0}dfrac{f(x)}x$ does not necessarily exist.
$endgroup$
– Saad
yesterday
$begingroup$
@maxmilgram I make a difference quotient by force to make use of assumption that f is differentiable. $lim_{y rightarrow x} frac{h(x) - h(y)}{x-y}$ = $h'(x)$ where $h(x) = frac{f(x)}{x^2}$ Also I make use of standard limit $lim_{z rightarrow 0} frac{sin z}{z} = 1$. But
$endgroup$
– Bag of Chips
yesterday
$begingroup$
@maxmilgram I make a difference quotient by force to make use of assumption that f is differentiable. $lim_{y rightarrow x} frac{h(x) - h(y)}{x-y}$ = $h'(x)$ where $h(x) = frac{f(x)}{x^2}$ Also I make use of standard limit $lim_{z rightarrow 0} frac{sin z}{z} = 1$. But
$endgroup$
– Bag of Chips
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Applying l'Hospital's rule to the inner limit gives
$-(2xf(x) - x^2f'(x))/(1 + cos x)$.
Now applying the rule to the outer limit gives
$(2f(x) - x^2f''(x))/sin x$
which diverges for bounded f and f" unless f(0) = 0.
answered yesterday
William ElliotWilliam Elliot
8,5922720
$endgroup$
$begingroup$
Does the second derivative need to exist?
$endgroup$
– Bag of Chips
yesterday
$begingroup$
@BagofChips. I assumed it did to explore the problem.
$endgroup$
– William Elliot
yesterday
add a comment |
$begingroup$
Split the numerator as $$y^2f(x)-x^2f(x)+x^2f(x)-x^2f(y)$$ and replace the $sin(x-y) $ with $(x-y) $ to get the limit $yto x$ as $$frac{x^2f'(x)-2xf(x)}{1-cos x} $$ and the desired limit as $xto 0$ of above fraction exists if and only if limit $L=lim_{xto 0}f(x)/x$ exists and in that case the desired limit is $2f'(0)-4L$.
answered yesterday
Paramanand SinghParamanand Singh
50.6k557168
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Applying l'Hospital's rule to the inner limit gives
$-(2xf(x) - x^2f'(x))/(1 + cos x)$.
Now applying the rule to the outer limit gives
$(2f(x) - x^2f''(x))/sin x$
which diverges for bounded f and f" unless f(0) = 0.
answered yesterday
William ElliotWilliam Elliot
8,5922720
$endgroup$
$begingroup$
Does the second derivative need to exist?
$endgroup$
– Bag of Chips
yesterday
$begingroup$
@BagofChips. I assumed it did to explore the problem.
$endgroup$
– William Elliot
yesterday
add a comment |
$begingroup$
Applying l'Hospital's rule to the inner limit gives
$-(2xf(x) - x^2f'(x))/(1 + cos x)$.
Now applying the rule to the outer limit gives
$(2f(x) - x^2f''(x))/sin x$
which diverges for bounded f and f" unless f(0) = 0.
answered yesterday
William ElliotWilliam Elliot
8,5922720
$endgroup$
$begingroup$
Does the second derivative need to exist?
$endgroup$
– Bag of Chips
yesterday
$begingroup$
@BagofChips. I assumed it did to explore the problem.
$endgroup$
– William Elliot
yesterday
add a comment |
$begingroup$
Applying l'Hospital's rule to the inner limit gives
$-(2xf(x) - x^2f'(x))/(1 + cos x)$.
Now applying the rule to the outer limit gives
$(2f(x) - x^2f''(x))/sin x$
which diverges for bounded f and f" unless f(0) = 0.
answered yesterday
William ElliotWilliam Elliot
8,5922720
$endgroup$
Applying l'Hospital's rule to the inner limit gives
$-(2xf(x) - x^2f'(x))/(1 + cos x)$.
Now applying the rule to the outer limit gives
$(2f(x) - x^2f''(x))/sin x$
which diverges for bounded f and f" unless f(0) = 0.
answered yesterday
William ElliotWilliam Elliot
8,5922720
answered yesterday
William ElliotWilliam Elliot
8,5922720
answered yesterday
William ElliotWilliam Elliot
8,5922720
answered yesterday
William ElliotWilliam Elliot
8,5922720
8,5922720
$begingroup$
Does the second derivative need to exist?
$endgroup$
– Bag of Chips
yesterday
$begingroup$
@BagofChips. I assumed it did to explore the problem.
$endgroup$
– William Elliot
yesterday
add a comment |
$begingroup$
Does the second derivative need to exist?
$endgroup$
– Bag of Chips
yesterday
$begingroup$
@BagofChips. I assumed it did to explore the problem.
$endgroup$
– William Elliot
yesterday
$begingroup$
Does the second derivative need to exist?
$endgroup$
– Bag of Chips
yesterday
$begingroup$
Does the second derivative need to exist?
$endgroup$
– Bag of Chips
yesterday
$begingroup$
@BagofChips. I assumed it did to explore the problem.
$endgroup$
– William Elliot
yesterday
$begingroup$
@BagofChips. I assumed it did to explore the problem.
$endgroup$
– William Elliot
yesterday
add a comment |
$begingroup$
Split the numerator as $$y^2f(x)-x^2f(x)+x^2f(x)-x^2f(y)$$ and replace the $sin(x-y) $ with $(x-y) $ to get the limit $yto x$ as $$frac{x^2f'(x)-2xf(x)}{1-cos x} $$ and the desired limit as $xto 0$ of above fraction exists if and only if limit $L=lim_{xto 0}f(x)/x$ exists and in that case the desired limit is $2f'(0)-4L$.
answered yesterday
Paramanand SinghParamanand Singh
50.6k557168
$endgroup$
add a comment |
$begingroup$
Split the numerator as $$y^2f(x)-x^2f(x)+x^2f(x)-x^2f(y)$$ and replace the $sin(x-y) $ with $(x-y) $ to get the limit $yto x$ as $$frac{x^2f'(x)-2xf(x)}{1-cos x} $$ and the desired limit as $xto 0$ of above fraction exists if and only if limit $L=lim_{xto 0}f(x)/x$ exists and in that case the desired limit is $2f'(0)-4L$.
answered yesterday
Paramanand SinghParamanand Singh
50.6k557168
$endgroup$
add a comment |
$begingroup$
Split the numerator as $$y^2f(x)-x^2f(x)+x^2f(x)-x^2f(y)$$ and replace the $sin(x-y) $ with $(x-y) $ to get the limit $yto x$ as $$frac{x^2f'(x)-2xf(x)}{1-cos x} $$ and the desired limit as $xto 0$ of above fraction exists if and only if limit $L=lim_{xto 0}f(x)/x$ exists and in that case the desired limit is $2f'(0)-4L$.
answered yesterday
Paramanand SinghParamanand Singh
50.6k557168
$endgroup$
Split the numerator as $$y^2f(x)-x^2f(x)+x^2f(x)-x^2f(y)$$ and replace the $sin(x-y) $ with $(x-y) $ to get the limit $yto x$ as $$frac{x^2f'(x)-2xf(x)}{1-cos x} $$ and the desired limit as $xto 0$ of above fraction exists if and only if limit $L=lim_{xto 0}f(x)/x$ exists and in that case the desired limit is $2f'(0)-4L$.
answered yesterday
Paramanand SinghParamanand Singh
50.6k557168
answered yesterday
Paramanand SinghParamanand Singh
50.6k557168
answered yesterday
Paramanand SinghParamanand Singh
50.6k557168
answered yesterday
Paramanand SinghParamanand Singh
50.6k557168
50.6k557168
add a comment |
add a comment |
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$begingroup$
It seems that some condition like $f(0)=0$ is actually missing instead.
$endgroup$
– Saad
yesterday
$begingroup$
@Saad I forgot to add: derivative of f is continuous.
$endgroup$
– Bag of Chips
yesterday
$begingroup$
Please explain how you evaluate the inner limit. I get the same result but without the term involving $f(x)$.
$endgroup$
– maxmilgram
yesterday
1
$begingroup$
The power of $x$ in $dfrac x{1-cos x}simdfrac2x$ is indeed correct, but $limlimits_{x→0}dfrac{f(x)}x$ does not necessarily exist.
$endgroup$
– Saad
yesterday
$begingroup$
@maxmilgram I make a difference quotient by force to make use of assumption that f is differentiable. $lim_{y rightarrow x} frac{h(x) - h(y)}{x-y}$ = $h'(x)$ where $h(x) = frac{f(x)}{x^2}$ Also I make use of standard limit $lim_{z rightarrow 0} frac{sin z}{z} = 1$. But
$endgroup$
– Bag of Chips
yesterday