using infinity to use L'Hôpitals rule: lim[x->infty](3x^5-2x^3+4)/(3+x-4x^5)

[MPR]

using infinity to use L'Hôpitals rule: lim[x->infty](3x^5-2x^3+4)/(3+x-4x^5)

Goodday,

I am currently having problems solving the following limit:

. . . . .\(\displaystyle \large{\displaystyle \lim_{x \rightarrow \infty}\, \left(\dfrac{3x^5\, -\, 2x^3\, +\, 4}{3\, +\, x\, -\, 4x^5}\right)}\)

I'd like to solve this using L'Hôpitals rule. For this this to be possible the equation needs to equal \(\displaystyle \infty/\infty\), which it should.

However when I try to fill in the equation using \(\displaystyle \infty,\) Ifail to see why this would result in \(\displaystyle \infty/\infty\)

I have been taught that \(\displaystyle \infty\, +\, \infty\, =\, \infty,\) and \(\displaystyle \infty\, -\, \infty\) is undefined.

So I see why \(\displaystyle 3x^5\, -\, 2x^3\, +\, 4\, =\, \infty,\) but not why 3 + x - 4x^(5) = \(\displaystyle \infty.\)

​

If anything should this not be zero because \(\displaystyle 4\, \times\, \infty^5\) seems to me a bigger number than \(\displaystyle \infty,\) so that \(\displaystyle 3\, +\, \infty\, -\, 4\, \times\, \infty^5\, =\, 0\)

I know I am not suppose to think like this because "infinity" isn't a number, but still it does not make sense to me.

Thank you.

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[tkhunny]

Goodday,

I am currently having problems solving the following limit:

View attachment 9631

I'd like to solve this using L'Hôpitals rule. For this this to be possible the equation needs to equal

/

, which it should.
H

owever​

when I try to fill in the equation using

, I​

fail to see why this would result in

/

.

I have been thought that:

+

=

-

= undefined

So I see why 3x^(5) -2x^(3) + 4 =

​

, but not why 3 + x - 4x^(5) =

​

.
If anything should this not be zero because 4*

​

^(5) seems to me a bigger number than

, so that

3 +

​

- 4*

​

^(5) = 0.
I know I am not suppose to think like this because

​

isn't a number but still it does not make sense to me.

Thank you.

​

​

​

​

​

​

​

​

There is a reason why it does not make sense to you. Don't do that! It doesn't make sense to anyone when you do that.

If x = 0, what is it all about?

View attachment 9631

4/3 -- Notice how the constants are VERY IMPORTANT around x = 0.

How about x = 2?

(96 - 16 + 4)/(3 + 2 - 128) = 84/(-123) -- Notice how the constants faded away in importance? They gave way to the highest order terms.

How about x = 1000?

(3*ReallyBig - PrettyBig + 4)/(3 + 1000 - 4*ReallyBig) -- See how the highest order terms are just totally taking over?

Eventually, as we ponder x increasing without bound...

(3*HUGE +/- NoOneCares)/(NoOneCares - 4*HUGE) -- Do you see where this is going?

Note: It will help you think about it if you read this symbol: \(\displaystyle x\rightarrow\infty\) "as x increases without bound" rather than the easily misunderstood "as x approaches infinity".

 

[MPR]

There is a reason why it does not make sense to you. Don't do that! It doesn't make sense to anyone when you do that.

If x = 0, what is it all about?

View attachment 9631

4/3 -- Notice how the constants are VERY IMPORTANT around x = 0.

How about x = 2?

(96 - 16 + 4)/(3 + 2 - 128) = 84/(-123) -- Notice how the constants faded away in importance? They gave way to the highest order terms.

How about x = 1000?

(3*ReallyBig - PrettyBig + 4)/(3 + 1000 - 4*ReallyBig) -- See how the highest order terms are just totally taking over?

Eventually, as we ponder x increasing without bound...

(3*HUGE +/- NoOneCares)/(NoOneCares - 4*HUGE) -- Do you see where this is going?

Note: It will help you think about it if you read this symbol: \(\displaystyle x\rightarrow\infty\) "as x increases without bound" rather than the easily misunderstood "as x approaches infinity".

Thank you for the response,

Okay, so especially when dealing with infinity, the term with the highest power will determine the result of the equation.
all terms where the exponent is smaller sort of "fades". This indeed makes more sense.

Thinking like "x increases without bound" instead of "x aproaches infinity" does help. Thank you.