#13**+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**+15 **

d) lim x→3 [1/(x-3) - 9/(x^{2} - 9)] Notice that this simplifies to

lim x→3 [(x - 6)/(x^{2} - 9)]

Note that when

x → 3^{-} .... [x - 6)/(x^{2} - 9)] → ∞ and when x → 3^{+} [x - 6)/(x^{2} - 9)] → - ∞

Thus .......lim x→3 [x - 6)/(x^{2} - 9)] = d.n.e.

e) lim x →∞ [(3 - cosx)/ (x^{2} + 10)] .......note that we can write this as

lim x →∞ [(3 / (x^{2} + 10)] - lim x →∞ [( cosx)/ (x^{2} + 10)]

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

For the second, note that

-1 ≤ cosx ≤ 1 and dividing everything by x^{2} + 10, we have

[-1/x^{2} + 10] ≤ [cosx/x^{2} + 10] ≤ [1/x^{2} + 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)/ (x^{2} + 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→∞ e^{0} = 1

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

CPhill
Oct 30, 2014

#2**+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

#4**+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**+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

#7**+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**+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

#10**+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**+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

#13**+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

#15**+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**+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

#19**+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

#23**+10 **

Note xvxvxv.......here's the graph of sinx^{2} / x^{2} .........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