+0  
 
0
758
24
avatar+1832 

I want the answers for d,  e, f 

 

xvxvxv  Oct 30, 2014

Best Answer 

 #13
avatar+94183 
+15

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
 #1
avatar+92857 
+15

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
 #2
avatar+94183 
+15

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
 #3
avatar+92857 
+10

d.n.e. = does not exist

 

CPhill  Oct 30, 2014
 #4
avatar+94183 
+5

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.

 

But what about c?    

Melody  Oct 30, 2014
 #5
avatar+92857 
+5

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
 #6
avatar+27237 
+10

Here's a way of doing a) without using L'Hopital's rule:

limit

.

Alan  Oct 30, 2014
 #7
avatar+92857 
+5

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
 #8
avatar+94183 
+10

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
 #9
avatar+92857 
+5

Very nice !!!!....you're not as "limited" as you think !!!!

 

CPhill  Oct 31, 2014
 #10
avatar+94183 
+5

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
 #11
avatar+94183 
+10

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
 #12
avatar+1832 
+5

for e I do it like this 

 

 

 

 

and melody , I dont understand c 

xvxvxv  Oct 31, 2014
 #13
avatar+94183 
+15
Best Answer

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
avatar+1832 
+5

yes thank you 

xvxvxv  Oct 31, 2014
 #15
avatar+94183 
+10

Your answer for e is good 315.

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
 #16
avatar+92857 
+10

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
 #17
avatar+1832 
0

thank you melody for your comment 

xvxvxv  Oct 31, 2014
 #18
avatar+1832 
0

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

xvxvxv  Oct 31, 2014
 #19
avatar+94183 
+10

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
 #20
avatar+1832 
0

thank you melody 

xvxvxv  Oct 31, 2014
 #21
avatar+1832 
0

CPhill 

 

 

for c   I do it like this 

 

xvxvxv  Oct 31, 2014
 #22
avatar+1832 
0

what about the last picture ^^^^^

xvxvxv  Nov 1, 2014
 #23
avatar+92857 
+10

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
 #24
avatar+1832 
0

Ok thank you 

xvxvxv  Nov 1, 2014

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