Limits question -a challenge (1 Viewer)

HeroicPandas · Nov 2, 2012

$\lim_{x \to\infty }\frac{x^2sin(\frac{1}{x})}{1+x}$

NO L HOPITAL RULE WHATEVER ITS CALLED, wolfram alpha used it...]

Can anyone look at my method to see if it is right, if there is another method, please post it up

My Method:

$=\lim_{x \to\infty }\frac{x^2sin(\frac{1}{x})}{1+x}$

$=\lim_{x \to\infty }\frac{x^2}{1+x}.\lim_{x \to\infty }sin\frac{1}{x}$

$=\lim_{x \to\infty }\frac{x}{1+x}.\lim_{x \to\infty }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

$=\lim_{x \to\infty }\frac{1}{\frac{1}{x}+1}.\lim_{x \to\infty }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

I will have to work slowly for the first limit just in-case there is confusion

$=\frac{1}{\frac{1}{\infty }+1}.\lim_{x \to\infty }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

$=\lim_{x \to\infty }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

NOW THIS IS THE PART... I reciprocated thingy under the limit

$=\lim_{\frac{1}{x} \to\frac{1}{\infty } }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

$=\lim_{\frac{1}{x} \to0 }\frac{sin\frac{1}{x}}{\frac{1}{x}}$
$= 1$

I would to know if there is another method

Shadowdude · Nov 2, 2012

just skimming

but why don't you make the change of variable u = 1/x and let u go to 0 instead

that should work.. i think. maybe. probably.

HeroicPandas · Nov 2, 2012

Shadowdude said:
just skimming

but why don't you make the change of variable u = 1/x and let u go to 0 instead

that should work.. i think. maybe. probably.

My teacher did that, but does my method work?

Shadowdude · Nov 2, 2012

You pretty much did the u-substitution method... without the u.

But it looks really suss with the 1/infinity

pokka · Nov 2, 2012

HeroicPandas said:
My teacher did that, but does my method work?

It's basically the same method, it's just let's you recognise the limit easily by changing the variable. So yes I think your method works.

SpiralFlex · Nov 2, 2012

Yes it's valid. If both of the following limits exist:

$\lim_{x\rightarrow a}f(x)$

$\lim_{x\rightarrow a}g(x)$

Then the following is true.

$\lim_{x\rightarrow a}f(x)*g(x)=\lim_{x\rightarrow a}f(x)*\lim_{x\rightarrow a}g(x)$

So that's fine.

oh well · Nov 2, 2012

SpiralFlex said:
Yes it's valid. If both of the following limits exist:

$\lim_{x\rightarrow a}f(x)$

$\lim_{x\rightarrow a}g(x)$

Then the following is true.

$\lim_{x\rightarrow a}f(x)*g(x)=\lim_{x\rightarrow a}f(x)*\lim_{x\rightarrow a}g(x)$

So that's fine.

Isn't that L'Hopital's Rule?

Shadowdude · Nov 2, 2012

oh well said:
Isn't that L'Hopital's Rule?

No, L'Hopital has stuff to do with derivatives.

Not school stuff so I'll leave it to you to search for yourself.

oh well · Nov 2, 2012

HeroicPandas said:
$\lim_{x \to\infty }\frac{x^2sin(\frac{1}{x})}{1+x}$

NO L HOSPITAL RULE WHATEVER ITS CALLED, wolfram alpha used it...]

Can anyone look at my method to see if it is right, if there is another method, please post it up

My Method:

$=\lim_{x \to\infty }\frac{x^2sin(\frac{1}{x})}{1+x}$

$=\lim_{x \to\infty }\frac{x^2}{1+x}.\lim_{x \to\infty }sin\frac{1}{x}$

$=\lim_{x \to\infty }\frac{x}{1+x}.\lim_{x \to\infty }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

$=\lim_{x \to\infty }\frac{1}{\frac{1}{x}+1}.\lim_{x \to\infty }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

I will have to work slowly for the first limit just in-case there is confusion

$=\frac{1}{\frac{1}{\infty }+1}.\lim_{x \to\infty }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

$=\lim_{x \to\infty }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

NOW THIS IS THE PART... I reciprocated thingy under the limit

$=\lim_{\frac{1}{x} \to\frac{1}{\infty } }\frac{sin\frac{1}{x}}{\frac{1}{x}}$

$=\lim_{\frac{1}{x} \to0 }\frac{sin\frac{1}{x}}{\frac{1}{x}}$
$= 1$

I would to know if there is another method

It's L'Hopital, not HOSPITAL LMAO

SpiralFlex · Nov 2, 2012

oh well said:
Isn't that L'Hospital's Rule?

No. L'Hopital's rule is different.

$Let functions f and g are differentiable on the open interval$\;(a,b)\;$containing$\; c\;$except itself and also the limit of$\; \frac{f(x)}{g(x)}\;$as$\;x\;$approaches$\; c\; $produces an indeterminate form then$

$\lim_{x\rightarrow c}\frac{f(x)}{g(x)}=\lim_{x\rightarrow c}\frac{f'(x)}{g'(x)}$

Edit: Sorry I forgot this was 3U. Never mind.

HeroicPandas · Nov 2, 2012

SpiralFlex said:
Yes it's valid. If both of the following limits exist:

$\lim_{x\rightarrow a}f(x)$

$\lim_{x\rightarrow a}g(x)$

Then the following is true.

$\lim_{x\rightarrow a}f(x)*g(x)=\lim_{x\rightarrow a}f(x)*\lim_{x\rightarrow a}g(x)$

So that's fine.

ok thank you

seanieg89 · Nov 2, 2012

oh well said:
It's L'Hopital, not HOSPITAL LMAO

There are multiple spellings.

Carrotsticks · Nov 2, 2012

Limits question -a challenge (1 Viewer)

HeroicPandas

Heroic!

Shadowdude

Cult of Personality

HeroicPandas

Heroic!

Shadowdude

Cult of Personality

pokka

d'ohnuts

SpiralFlex

Well-Known Member

oh well

Banned

Shadowdude

Cult of Personality

oh well

Banned

SpiralFlex

Well-Known Member

HeroicPandas

Heroic!

seanieg89

Well-Known Member

Carrotsticks

Retired

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