Limit with sandwich theorem: $limlimits_{xto 0}left(xsin(1/x)+xcos(x)right).$












0












$begingroup$



Evaluate
$$lim_{xto 0}left(xsin(1/x)+xcos(x)right).$$




I don't know how to solve it and I have just one thought in my mind, maybe I can solve it with sandwich theorem but I don't know why, can you tell its cause and solve this question?






calculus limits







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Hint. Note that $sin$ and $cos$ values belong to $[-1,1]$.
    $endgroup$
    – Robert Z
    Dec 4 '18 at 6:29










  • $begingroup$
    You only need sandwich on first term of the sum, since second term goes to zero.
    $endgroup$
    – coffeemath
    Dec 4 '18 at 6:29










  • $begingroup$
    So how can we apply to first term to sandwich theorem?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:30










  • $begingroup$
    You can find a few posts on this site about the first term using Approach0. For example: $lim x sin (1/x)$, when $x to 0$.
    $endgroup$
    – Martin Sleziak
    Dec 5 '18 at 12:49










  • $begingroup$
    @MartinSleziak thank you
    $endgroup$
    – Atakan
    Dec 5 '18 at 13:02
















0












$begingroup$



Evaluate
$$lim_{xto 0}left(xsin(1/x)+xcos(x)right).$$




I don't know how to solve it and I have just one thought in my mind, maybe I can solve it with sandwich theorem but I don't know why, can you tell its cause and solve this question?






calculus limits







share|cite|improve this question











$endgroup$








  • 6




    $begingroup$
    Hint. Note that $sin$ and $cos$ values belong to $[-1,1]$.
    $endgroup$
    – Robert Z
    Dec 4 '18 at 6:29










  • $begingroup$
    You only need sandwich on first term of the sum, since second term goes to zero.
    $endgroup$
    – coffeemath
    Dec 4 '18 at 6:29










  • $begingroup$
    So how can we apply to first term to sandwich theorem?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:30










  • $begingroup$
    You can find a few posts on this site about the first term using Approach0. For example: $lim x sin (1/x)$, when $x to 0$.
    $endgroup$
    – Martin Sleziak
    Dec 5 '18 at 12:49










  • $begingroup$
    @MartinSleziak thank you
    $endgroup$
    – Atakan
    Dec 5 '18 at 13:02














0












0








0





$begingroup$



Evaluate
$$lim_{xto 0}left(xsin(1/x)+xcos(x)right).$$




I don't know how to solve it and I have just one thought in my mind, maybe I can solve it with sandwich theorem but I don't know why, can you tell its cause and solve this question?






calculus limits







share|cite|improve this question











$endgroup$





Evaluate
$$lim_{xto 0}left(xsin(1/x)+xcos(x)right).$$




I don't know how to solve it and I have just one thought in my mind, maybe I can solve it with sandwich theorem but I don't know why, can you tell its cause and solve this question?






calculus limits




calculus limits






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 12:47









Martin Sleziak

44.7k9117272




44.7k9117272










asked Dec 4 '18 at 6:24









AtakanAtakan

184




184








  • 6




    $begingroup$
    Hint. Note that $sin$ and $cos$ values belong to $[-1,1]$.
    $endgroup$
    – Robert Z
    Dec 4 '18 at 6:29










  • $begingroup$
    You only need sandwich on first term of the sum, since second term goes to zero.
    $endgroup$
    – coffeemath
    Dec 4 '18 at 6:29










  • $begingroup$
    So how can we apply to first term to sandwich theorem?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:30










  • $begingroup$
    You can find a few posts on this site about the first term using Approach0. For example: $lim x sin (1/x)$, when $x to 0$.
    $endgroup$
    – Martin Sleziak
    Dec 5 '18 at 12:49










  • $begingroup$
    @MartinSleziak thank you
    $endgroup$
    – Atakan
    Dec 5 '18 at 13:02














  • 6




    $begingroup$
    Hint. Note that $sin$ and $cos$ values belong to $[-1,1]$.
    $endgroup$
    – Robert Z
    Dec 4 '18 at 6:29










  • $begingroup$
    You only need sandwich on first term of the sum, since second term goes to zero.
    $endgroup$
    – coffeemath
    Dec 4 '18 at 6:29










  • $begingroup$
    So how can we apply to first term to sandwich theorem?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:30










  • $begingroup$
    You can find a few posts on this site about the first term using Approach0. For example: $lim x sin (1/x)$, when $x to 0$.
    $endgroup$
    – Martin Sleziak
    Dec 5 '18 at 12:49










  • $begingroup$
    @MartinSleziak thank you
    $endgroup$
    – Atakan
    Dec 5 '18 at 13:02








6




6




$begingroup$
Hint. Note that $sin$ and $cos$ values belong to $[-1,1]$.
$endgroup$
– Robert Z
Dec 4 '18 at 6:29




$begingroup$
Hint. Note that $sin$ and $cos$ values belong to $[-1,1]$.
$endgroup$
– Robert Z
Dec 4 '18 at 6:29












$begingroup$
You only need sandwich on first term of the sum, since second term goes to zero.
$endgroup$
– coffeemath
Dec 4 '18 at 6:29




$begingroup$
You only need sandwich on first term of the sum, since second term goes to zero.
$endgroup$
– coffeemath
Dec 4 '18 at 6:29












$begingroup$
So how can we apply to first term to sandwich theorem?
$endgroup$
– Atakan
Dec 4 '18 at 6:30




$begingroup$
So how can we apply to first term to sandwich theorem?
$endgroup$
– Atakan
Dec 4 '18 at 6:30












$begingroup$
You can find a few posts on this site about the first term using Approach0. For example: $lim x sin (1/x)$, when $x to 0$.
$endgroup$
– Martin Sleziak
Dec 5 '18 at 12:49




$begingroup$
You can find a few posts on this site about the first term using Approach0. For example: $lim x sin (1/x)$, when $x to 0$.
$endgroup$
– Martin Sleziak
Dec 5 '18 at 12:49












$begingroup$
@MartinSleziak thank you
$endgroup$
– Atakan
Dec 5 '18 at 13:02




$begingroup$
@MartinSleziak thank you
$endgroup$
– Atakan
Dec 5 '18 at 13:02










3 Answers
3






active

oldest

votes


















1












$begingroup$

It follows from $-2|x|le x(sinfrac{1}{x}+cos x)le 2|x|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why do we apply sandwich theorem to x.cosx? Also why are they between -2|x| and 2|x|? Can you explain it clearly?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:47






  • 1




    $begingroup$
    @Atakan Treating the terms separately is optional, since $lim_{xto 0}cos x$ exists, but I decided to make the entire proof succinct. The limits follow from $sinfrac{1}{x},,cos xin [-1,,1]impliessinfrac{1}{x}+cos xin [-2,,2]$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:50



















2












$begingroup$

Hint



Note, that for every $ alpha, beta in mathbb{R}$ we have
$$|sin alpha + cos beta| leq 2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where does this formula come from?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:48










  • $begingroup$
    @Atakan From the triangle inequality.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:51










  • $begingroup$
    @J.G. Yes I know this but I care why we are using this,this is important for me
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:52










  • $begingroup$
    @Atakan Because if $fto 0$ and $g$ is bounded, $fgto 0$ by the sandwich theorem, and in this case we can take $f=x,,g=sinfrac{1}{x}+cos x$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:56










  • $begingroup$
    @J.G. thank you
    $endgroup$
    – Atakan
    Dec 4 '18 at 7:00



















1












$begingroup$

Both $sin(x)$ and $cos(x)$ can have values in the range [-1,1].



Therefore:



$-2x=(-1)x+(-1)xleq xsin(frac1x)+xcos xleq x+x=2x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for your clear explanation
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:57










  • $begingroup$
    Having a look at "J.G."s answer I see that I made a mistake: I assumed $lim_{xto 0+}$ instead of $lim_{xto 0}$ so I wrote $x$ instead of $|x|$.
    $endgroup$
    – Martin Rosenau
    Dec 4 '18 at 11:31










  • $begingroup$
    no problem I get it what you mean
    $endgroup$
    – Atakan
    Dec 4 '18 at 16:06











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

It follows from $-2|x|le x(sinfrac{1}{x}+cos x)le 2|x|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why do we apply sandwich theorem to x.cosx? Also why are they between -2|x| and 2|x|? Can you explain it clearly?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:47






  • 1




    $begingroup$
    @Atakan Treating the terms separately is optional, since $lim_{xto 0}cos x$ exists, but I decided to make the entire proof succinct. The limits follow from $sinfrac{1}{x},,cos xin [-1,,1]impliessinfrac{1}{x}+cos xin [-2,,2]$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:50
















1












$begingroup$

It follows from $-2|x|le x(sinfrac{1}{x}+cos x)le 2|x|$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Why do we apply sandwich theorem to x.cosx? Also why are they between -2|x| and 2|x|? Can you explain it clearly?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:47






  • 1




    $begingroup$
    @Atakan Treating the terms separately is optional, since $lim_{xto 0}cos x$ exists, but I decided to make the entire proof succinct. The limits follow from $sinfrac{1}{x},,cos xin [-1,,1]impliessinfrac{1}{x}+cos xin [-2,,2]$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:50














1












1








1





$begingroup$

It follows from $-2|x|le x(sinfrac{1}{x}+cos x)le 2|x|$.






share|cite|improve this answer









$endgroup$



It follows from $-2|x|le x(sinfrac{1}{x}+cos x)le 2|x|$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 6:33









J.G.J.G.

24.9k22539




24.9k22539












  • $begingroup$
    Why do we apply sandwich theorem to x.cosx? Also why are they between -2|x| and 2|x|? Can you explain it clearly?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:47






  • 1




    $begingroup$
    @Atakan Treating the terms separately is optional, since $lim_{xto 0}cos x$ exists, but I decided to make the entire proof succinct. The limits follow from $sinfrac{1}{x},,cos xin [-1,,1]impliessinfrac{1}{x}+cos xin [-2,,2]$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:50


















  • $begingroup$
    Why do we apply sandwich theorem to x.cosx? Also why are they between -2|x| and 2|x|? Can you explain it clearly?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:47






  • 1




    $begingroup$
    @Atakan Treating the terms separately is optional, since $lim_{xto 0}cos x$ exists, but I decided to make the entire proof succinct. The limits follow from $sinfrac{1}{x},,cos xin [-1,,1]impliessinfrac{1}{x}+cos xin [-2,,2]$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:50
















$begingroup$
Why do we apply sandwich theorem to x.cosx? Also why are they between -2|x| and 2|x|? Can you explain it clearly?
$endgroup$
– Atakan
Dec 4 '18 at 6:47




$begingroup$
Why do we apply sandwich theorem to x.cosx? Also why are they between -2|x| and 2|x|? Can you explain it clearly?
$endgroup$
– Atakan
Dec 4 '18 at 6:47




1




1




$begingroup$
@Atakan Treating the terms separately is optional, since $lim_{xto 0}cos x$ exists, but I decided to make the entire proof succinct. The limits follow from $sinfrac{1}{x},,cos xin [-1,,1]impliessinfrac{1}{x}+cos xin [-2,,2]$.
$endgroup$
– J.G.
Dec 4 '18 at 6:50




$begingroup$
@Atakan Treating the terms separately is optional, since $lim_{xto 0}cos x$ exists, but I decided to make the entire proof succinct. The limits follow from $sinfrac{1}{x},,cos xin [-1,,1]impliessinfrac{1}{x}+cos xin [-2,,2]$.
$endgroup$
– J.G.
Dec 4 '18 at 6:50











2












$begingroup$

Hint



Note, that for every $ alpha, beta in mathbb{R}$ we have
$$|sin alpha + cos beta| leq 2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where does this formula come from?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:48










  • $begingroup$
    @Atakan From the triangle inequality.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:51










  • $begingroup$
    @J.G. Yes I know this but I care why we are using this,this is important for me
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:52










  • $begingroup$
    @Atakan Because if $fto 0$ and $g$ is bounded, $fgto 0$ by the sandwich theorem, and in this case we can take $f=x,,g=sinfrac{1}{x}+cos x$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:56










  • $begingroup$
    @J.G. thank you
    $endgroup$
    – Atakan
    Dec 4 '18 at 7:00
















2












$begingroup$

Hint



Note, that for every $ alpha, beta in mathbb{R}$ we have
$$|sin alpha + cos beta| leq 2$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Where does this formula come from?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:48










  • $begingroup$
    @Atakan From the triangle inequality.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:51










  • $begingroup$
    @J.G. Yes I know this but I care why we are using this,this is important for me
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:52










  • $begingroup$
    @Atakan Because if $fto 0$ and $g$ is bounded, $fgto 0$ by the sandwich theorem, and in this case we can take $f=x,,g=sinfrac{1}{x}+cos x$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:56










  • $begingroup$
    @J.G. thank you
    $endgroup$
    – Atakan
    Dec 4 '18 at 7:00














2












2








2





$begingroup$

Hint



Note, that for every $ alpha, beta in mathbb{R}$ we have
$$|sin alpha + cos beta| leq 2$$






share|cite|improve this answer









$endgroup$



Hint



Note, that for every $ alpha, beta in mathbb{R}$ we have
$$|sin alpha + cos beta| leq 2$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 6:36









Jaroslaw MatlakJaroslaw Matlak

4,181930




4,181930












  • $begingroup$
    Where does this formula come from?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:48










  • $begingroup$
    @Atakan From the triangle inequality.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:51










  • $begingroup$
    @J.G. Yes I know this but I care why we are using this,this is important for me
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:52










  • $begingroup$
    @Atakan Because if $fto 0$ and $g$ is bounded, $fgto 0$ by the sandwich theorem, and in this case we can take $f=x,,g=sinfrac{1}{x}+cos x$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:56










  • $begingroup$
    @J.G. thank you
    $endgroup$
    – Atakan
    Dec 4 '18 at 7:00


















  • $begingroup$
    Where does this formula come from?
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:48










  • $begingroup$
    @Atakan From the triangle inequality.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:51










  • $begingroup$
    @J.G. Yes I know this but I care why we are using this,this is important for me
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:52










  • $begingroup$
    @Atakan Because if $fto 0$ and $g$ is bounded, $fgto 0$ by the sandwich theorem, and in this case we can take $f=x,,g=sinfrac{1}{x}+cos x$.
    $endgroup$
    – J.G.
    Dec 4 '18 at 6:56










  • $begingroup$
    @J.G. thank you
    $endgroup$
    – Atakan
    Dec 4 '18 at 7:00
















$begingroup$
Where does this formula come from?
$endgroup$
– Atakan
Dec 4 '18 at 6:48




$begingroup$
Where does this formula come from?
$endgroup$
– Atakan
Dec 4 '18 at 6:48












$begingroup$
@Atakan From the triangle inequality.
$endgroup$
– J.G.
Dec 4 '18 at 6:51




$begingroup$
@Atakan From the triangle inequality.
$endgroup$
– J.G.
Dec 4 '18 at 6:51












$begingroup$
@J.G. Yes I know this but I care why we are using this,this is important for me
$endgroup$
– Atakan
Dec 4 '18 at 6:52




$begingroup$
@J.G. Yes I know this but I care why we are using this,this is important for me
$endgroup$
– Atakan
Dec 4 '18 at 6:52












$begingroup$
@Atakan Because if $fto 0$ and $g$ is bounded, $fgto 0$ by the sandwich theorem, and in this case we can take $f=x,,g=sinfrac{1}{x}+cos x$.
$endgroup$
– J.G.
Dec 4 '18 at 6:56




$begingroup$
@Atakan Because if $fto 0$ and $g$ is bounded, $fgto 0$ by the sandwich theorem, and in this case we can take $f=x,,g=sinfrac{1}{x}+cos x$.
$endgroup$
– J.G.
Dec 4 '18 at 6:56












$begingroup$
@J.G. thank you
$endgroup$
– Atakan
Dec 4 '18 at 7:00




$begingroup$
@J.G. thank you
$endgroup$
– Atakan
Dec 4 '18 at 7:00











1












$begingroup$

Both $sin(x)$ and $cos(x)$ can have values in the range [-1,1].



Therefore:



$-2x=(-1)x+(-1)xleq xsin(frac1x)+xcos xleq x+x=2x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for your clear explanation
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:57










  • $begingroup$
    Having a look at "J.G."s answer I see that I made a mistake: I assumed $lim_{xto 0+}$ instead of $lim_{xto 0}$ so I wrote $x$ instead of $|x|$.
    $endgroup$
    – Martin Rosenau
    Dec 4 '18 at 11:31










  • $begingroup$
    no problem I get it what you mean
    $endgroup$
    – Atakan
    Dec 4 '18 at 16:06
















1












$begingroup$

Both $sin(x)$ and $cos(x)$ can have values in the range [-1,1].



Therefore:



$-2x=(-1)x+(-1)xleq xsin(frac1x)+xcos xleq x+x=2x$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    thank you for your clear explanation
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:57










  • $begingroup$
    Having a look at "J.G."s answer I see that I made a mistake: I assumed $lim_{xto 0+}$ instead of $lim_{xto 0}$ so I wrote $x$ instead of $|x|$.
    $endgroup$
    – Martin Rosenau
    Dec 4 '18 at 11:31










  • $begingroup$
    no problem I get it what you mean
    $endgroup$
    – Atakan
    Dec 4 '18 at 16:06














1












1








1





$begingroup$

Both $sin(x)$ and $cos(x)$ can have values in the range [-1,1].



Therefore:



$-2x=(-1)x+(-1)xleq xsin(frac1x)+xcos xleq x+x=2x$






share|cite|improve this answer









$endgroup$



Both $sin(x)$ and $cos(x)$ can have values in the range [-1,1].



Therefore:



$-2x=(-1)x+(-1)xleq xsin(frac1x)+xcos xleq x+x=2x$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 6:52









Martin RosenauMartin Rosenau

1,156139




1,156139












  • $begingroup$
    thank you for your clear explanation
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:57










  • $begingroup$
    Having a look at "J.G."s answer I see that I made a mistake: I assumed $lim_{xto 0+}$ instead of $lim_{xto 0}$ so I wrote $x$ instead of $|x|$.
    $endgroup$
    – Martin Rosenau
    Dec 4 '18 at 11:31










  • $begingroup$
    no problem I get it what you mean
    $endgroup$
    – Atakan
    Dec 4 '18 at 16:06


















  • $begingroup$
    thank you for your clear explanation
    $endgroup$
    – Atakan
    Dec 4 '18 at 6:57










  • $begingroup$
    Having a look at "J.G."s answer I see that I made a mistake: I assumed $lim_{xto 0+}$ instead of $lim_{xto 0}$ so I wrote $x$ instead of $|x|$.
    $endgroup$
    – Martin Rosenau
    Dec 4 '18 at 11:31










  • $begingroup$
    no problem I get it what you mean
    $endgroup$
    – Atakan
    Dec 4 '18 at 16:06
















$begingroup$
thank you for your clear explanation
$endgroup$
– Atakan
Dec 4 '18 at 6:57




$begingroup$
thank you for your clear explanation
$endgroup$
– Atakan
Dec 4 '18 at 6:57












$begingroup$
Having a look at "J.G."s answer I see that I made a mistake: I assumed $lim_{xto 0+}$ instead of $lim_{xto 0}$ so I wrote $x$ instead of $|x|$.
$endgroup$
– Martin Rosenau
Dec 4 '18 at 11:31




$begingroup$
Having a look at "J.G."s answer I see that I made a mistake: I assumed $lim_{xto 0+}$ instead of $lim_{xto 0}$ so I wrote $x$ instead of $|x|$.
$endgroup$
– Martin Rosenau
Dec 4 '18 at 11:31












$begingroup$
no problem I get it what you mean
$endgroup$
– Atakan
Dec 4 '18 at 16:06




$begingroup$
no problem I get it what you mean
$endgroup$
– Atakan
Dec 4 '18 at 16:06


















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