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# limits

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I want the answers for d,  e, f

xvxvxv  Oct 30, 2014

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Hi 315

c)

Consider

lim x tends to infinity of   (some little positive number divided by x)

=a little positive number divided by a huge number

the limit will be 0.

like

9/1000000000000000000000000000000000000000000000000000000000

is a very tiny number and the more digits you add to the end of it the smaller and closer to zero it becomes.

Does that make sense?

Melody  Oct 31, 2014
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d) lim x→3 [1/(x-3) - 9/(x2 - 9)]  Notice that this simplifies to

lim x→3 [(x - 6)/(x2 - 9)]

Note that when

x → 3- .... [x - 6)/(x2 - 9)] →  ∞   and when x → 3+  [x - 6)/(x2 - 9)] →  - ∞

Thus .......lim x→3 [x - 6)/(x2 - 9)]   =  d.n.e.

e)  lim x →∞ [(3 - cosx)/ (x2 + 10)] .......note that we can write this as

lim x →∞ [(3 / (x2 + 10)] - lim x →∞ [( cosx)/ (x2 + 10)]

The first is easy.... it just evaluates to 0

For the second, note that

-1 ≤ cosx ≤ 1      and dividing everything by x2 + 10, we have

[-1/x2 + 10] ≤ [cosx/x2 + 10]  ≤ [1/x2 + 10]

Note that, as the two outside functions appproach infinity, they both approach 0. So, the middle function is "Squeezed" and it approaches 0, as well.

Therefore......  lim x →∞ [(3 - cosx)/ (x2 + 10)]  =  0

f) lim x→∞ e [(x+ 1)/(2x^2 + 1)]

This one is easy....note that.... lim x→∞ [(x+ 1)/(2x^2 + 1)] = 0

So ....lim x→∞ e [(x+ 1)/(2x^2 + 1)]  =  lim x→∞ e0  = 1

I'm not great at limits......but  I think these might be correct.....

CPhill  Oct 30, 2014
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okay.

I am really bad at limits.  I'll state that up front.

d)   I agree that d is undefined thaough I do not know what d.n.e stands for.

f)   I get the same as Chris too.

$$\displaystyle\lim_{x\rightarrow\infty}\;\;\frac{x+1}{2x^2+1}\\\\ =\displaystyle\lim_{x\rightarrow\infty}\;\;\frac{1}{4x+1}\\\\ =0\\\\\\ =\displaystyle\lim_{x\rightarrow\infty}\;\;e^0\\\\ =1$$

e) I'll admit to learning this one straight from CPhill.  Thanks Chris

$$\displaystyle\lim_{x\rightarrow\infty}\;\frac{3-cosx}{x^2+10}}\\\\ =\displaystyle\lim_{x\rightarrow\infty}\;\frac{3}{x^2+10}}\;\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\ =\;0\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\ =\;-\;\;\displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}}\\\\ Now -1\le cosx\le 1\\\\ So \\\\ \displaystyle\lim_{x\rightarrow\infty}\;\frac{-1}{x^2+10}\;\;\le \displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;\le\displaystyle\lim_{x\rightarrow\infty}\;\frac{1}{x^2+10}\\\\ 0\;\;\le \displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;\le\;0\\\\ therefore\\\\ \displaystyle\lim_{x\rightarrow\infty}\;\frac{cosx}{x^2+10}\;=\;0\\\\$$

$$\\therefore\\\\ \displaystyle\lim_{x\rightarrow\infty}\;\frac{3-cosx}{x^2+10}}=0-0=0\\\\$$

Thank you for showing me that Chris       (This answer is exactly the same as Chris's)

Melody  Oct 30, 2014
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d.n.e. = does not exist

CPhill  Oct 30, 2014
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More questions -  315 you can do these so perhaps you can tell me how please.

Otherwise maybe Chris or Alan would be able to assist please.

b)  I think I can do this one

If it approaches from above then the limit will be 1

Ig it approaches from below then the limit will be -1

therefore the limit is undefined.

I just worked out a.   you have to multiply top and bottom by the conjugate of the top and then it just falls out.

Melody  Oct 30, 2014
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Melody, I only know how to do (a) using L'Hospital's

Taking the derivative of the top and the bottom =

(1/2)(2)/√(2x + 4) = 1/√(2x + 4)

So

lim x → 0   1/√(2x + 4)  = 1/2

For (c)....note that     -1 ≤ sin x^2 ≤ 1    ....and I believe we can do this one like (e) using the Squeeze Theorem

CPhill  Oct 30, 2014
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Here's a way of doing a) without using L'Hopital's rule:

.

Alan  Oct 30, 2014
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Thanks for that one, Alan.....

Is the fact that we were taking the limit around 0 is what allowed us to use that Maclaurin expansion???? [I think that's what I remember from Cal 2 !!!!]

CPhill  Oct 30, 2014
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I repeat!

I just worked out a.   you have to multiply top and bottom by the conjugate of the top and then it just falls out.

It is quite simple this way.  :)

$$\displaystyle\lim_{x\rightarrow 0} \frac{\sqrt{2x+4}-2}{x}\\\\ =\displaystyle\lim_{x\rightarrow 0} \frac{\sqrt{2x+4}-2}{x}\times \frac{\sqrt{2x+4}+2}{\sqrt{2x+4}+2}\\\\ =\displaystyle\lim_{x\rightarrow 0} \frac{(2x+4)-4}{x(\sqrt{2x+4}+2)}\\\\ =\displaystyle\lim_{x\rightarrow 0} \frac{2}{(\sqrt{2x+4}+2)}\\\\ =\frac{2}{(\sqrt{2*0+4}+2)}\\\\ =\frac{2}{4}\\\\ =\frac{1}{2}$$

Melody  Oct 31, 2014
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Very nice !!!!....you're not as "limited" as you think !!!!

CPhill  Oct 31, 2014
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Thank you Chris,

I haven't finished with this question yet but I have put it in the Sticky Thread "Great answers to Learn from"  :)

Melody  Oct 31, 2014
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c)

$$\\\displaystyle\lim_{x\rightarrow \infty}\;\;\frac{sin(x^2)+10}{x^2+10}\\\\ Now\\ -1\le sin(x^2)\le 1\\ 9\le sin(x^2)+10\le 11\\ Therefore the numerator is a positive number between 9 and 11\\ Therefore\\ \\\displaystyle\lim_{x\rightarrow \infty}\;\;\frac{sin(x^2)+10}{x^2+10}=0$$

Melody  Oct 31, 2014
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for e I do it like this

and melody , I dont understand c

xvxvxv  Oct 31, 2014
#13
+91972
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Hi 315

c)

Consider

lim x tends to infinity of   (some little positive number divided by x)

=a little positive number divided by a huge number

the limit will be 0.

like

9/1000000000000000000000000000000000000000000000000000000000

is a very tiny number and the more digits you add to the end of it the smaller and closer to zero it becomes.

Does that make sense?

Melody  Oct 31, 2014
#14
+1832
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yes thank you

xvxvxv  Oct 31, 2014
#15
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My only criticism (and it may be just a person dislike) is this

If -3<x<4

then technically

3>-x>-4

this is true but I don't like the presentation.  I think the little  number should go first and that it should be rearranged and presented as

-4<-x<3

Maybe this is just a personal bias, another mathermatician might like to comment.

Melody  Oct 31, 2014
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Here's my approach to (c)

lim x → ∞  [sinx^2 + 10] /[x^ + 10]

Splitting this up, we have

lim x → ∞  [sinx^2 ] / [x^ + 10]  + lim x → ∞  [ 10] / [x^ + 10]

Notice that the second thing just evaluates to 0

For the first...we can do the Squeeze Theorem "thing" again

-1 ≤ sin x^2 ≤ 1

And dividing each thing by x^2 + 10, we have

-1 / x^2 + 10 ≤ sin x^2 / x^2 + 10  ≤ 1 / X^2 + 10

And because the two "outside" functions approach 0 as x approaches infinity, then the middle one is "squeezed" to this limit, too.

Thus

lim x → ∞  [sinx^2 ] / [x^ + 10]  + lim x → ∞  [ 10] / [x^ + 10]  =

0                              +               0   =                                0         !!!!

CPhill  Oct 31, 2014
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thank you melody for your comment

xvxvxv  Oct 31, 2014
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nice , but my question why do you put the angle ( x^2) not (x^2+10 )

xvxvxv  Oct 31, 2014
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Because

$$\\sin\;x^2+10\\ means \\(sin(x^2))+10\\ If you want the sine of  (x^2+10)\\ then it should be written as  sin(x^2+10)$$

Melody  Oct 31, 2014
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thank you melody

xvxvxv  Oct 31, 2014
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for c   I do it like this

xvxvxv  Oct 31, 2014
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what about the last picture ^^^^^

xvxvxv  Nov 1, 2014
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Note xvxvxv.......here's the graph of    sinx2 / x2   .........https://www.desmos.com/calculator/nflbtaael9

Note that the limit appears to be approaching 0 (not, 1) as x approaches infinity.....!!!!!

See my earlier answer on this one where I used the Squeeze Theorem......

CPhill  Nov 1, 2014
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Ok thank you

xvxvxv  Nov 1, 2014

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