Question

Can you please show me how to solve this?I would really appreciete it! (Limits and series)

http://tinypic.com/r/dy3x4m/8

Answer 1

\[\lim_{x\rightarrow -1} \frac{x^2+3x+2}{2x^2-8} \\ \lim_{x\rightarrow -2} \frac{x^2+3x+2}{2x^2-8}\] \[\lim_{x\rightarrow 1} \frac{\sqrt{1+3x}-2}{x-1}\] \[\lim_{(x~,y)\rightarrow (0,0)}\frac{x^2}{\sqrt{x^2+y^2}} \\ \lim_{(x~,~y)\rightarrow (0,0)} \frac{xy}{\sqrt{x^2+y^2}}\]

Answer 2

Ready for a race? @zzr0ck3r haha.

Answer 3

Heck no. I hate calculus :)

Answer 4

Okay in limit #1, if you directly plug in -1 for all the x's, what do you get, @xfire30 ?

Answer 5

We can apply L'Hospital's rule (if you're allowed to) in limit #2. This will become \(\sf \dfrac{2x+3}{4x}\) and plug in -2

Answer 6

6/-6 i get in limit #!?

Answer 7

is it 6/-6?

Answer 8

For limit #1?

Answer 9

yes

Answer 10

no. \((-1)^2+3(-1)+2 \ne 6\) but \(2(-1)^2-8 =-6\)

Answer 11

0/-6

Answer 12

Yes and what does that =..

plugging it into a calculator would be sooo simple.

Answer 13

0

Answer 14

good. limit #2, are you allowed to apply LH rule?

Answer 15

I am allowed

Answer 16

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Jhannybean
We can apply L'Hospital's rule (if you're allowed to) in limit #2. This will become \(\sf \dfrac{2x+3}{4x}\) and plug in -2
\(\color{#0cbb34}{\text{End of Quote}}\)

Answer 17

I dotn get it

Answer 18

How did he got (2x+3)/(4x)?

Answer 19

you get*

Answer 20

Ohhh,I get it now

Answer 21

When you apply L'Hospital's rule, you tage the derivative of the function in the numerator and the denominator.

Ok cool.

Answer 22

Derivated each term

Answer 23

The first 2 I understood,thank you so far :D

Answer 24

Yep, you derivitated each term. woot, np.

Answer 25

And now I plug in x=1?

Answer 26

How do I solve it?

Answer 27

Looks the same to me,except that little "+" and"-"

Answer 28

It's sort of like the squeeze theorem if they both approach the same number/ exists, then the two sided limit will also approach the same number as the one sided limits approach.

Answer 29

\[\lim_{x \rightarrow c} f(x) = L\]

then the limit approaching from left and right equal the same number L pretty much

Answer 30

I know it doesn't exist, but when I solve it it seems like it freaking does. Lol. Wth :(

Answer 31

It should exist

Answer 32

Mess around with it, apply L'Hopital's rule

Answer 33

Ok so no left / right hand limits.

Answer 34

wolfram gives 3/4? http://www.wolframalpha.com/input/?i=lim+x-%3E+1+%28%28%5Csqrt%281%2B3x%29-2%29%2F%28x-1%29%29

Answer 35

No there is left and right hand limits, and they equal, that's why it exists :P

Answer 36

Sounds about right

Answer 37

oh. dur... I forgot to apply chain rule.

Answer 38

\[\sf \frac{d}{dx} (\sqrt{1+3x}-2) =\frac{1}{2}(1+3x)^{- 1/2}(3)\] so you aplly LH rule I guess haha.

Answer 39

\[\sf \lim_{x\rightarrow 1} \frac{\sqrt{1+3x}-2}{x-1}\]Applying LH rule, you get: \[\lim_{x\rightarrow 1} \frac{\frac{1}{2}(1+3x)^{-1/2}(3)}{1}\]Now you just plug in 1 and you'll get your limit! Sorry about that earlier.

Answer 40

Dont worry,just give me a second to try to understand what you wrote :D

Answer 41

This one is hard

Answer 42

You just derivated each term,right?

Answer 43

yep

Answer 44

What's "d" and "dx"?

Answer 45

that's how you write the derivative, lol

Answer 46

\(\sf \frac{d}{dx}\) means im taking the derivative, \(\sf d\), with respect to x. \(\sf dx\)

Answer 47

(1+3x)^(−1/2)

Answer 48

how do i solve this part?

Answer 49

Yeah and don't forget the chain rule

Answer 50

4^(-1/2) how much is it?

Answer 51

No, no. Are you familiar with derivatives?

Answer 52

\[4^{\frac{ -1 }{ 2 }}\]

Answer 53

I just plugged X=1 in the result you got after derivating

Answer 54

Yeah, I just plugged it all into my calculator.

Answer 55

(1/2)* (1+ 3(1))^(-1/2) * (3)

Answer 56

(1+ 3(1))^(-1/2) this part I dont know how to calculate

Answer 57

You can also think of it as \(\dfrac{3}{2(1+3(1))^{1/2}}\)

Answer 58

You make a negative power positive by sending it to the nether realms, aka denominator

Answer 59

the fraction is the problem I dont know to solve as power

Answer 60

something^1/2

Answer 61

how is it done?

Answer 62

the 1/2 power is the square root.

Answer 63

Oh you mean plugging it into the calculator, or like physically solving it? \[\frac{3}{2\sqrt{1+3(1)}} \implies \frac{3}{2(1+3(1))^{1/2}}\]

Answer 64

physically solving it

Answer 65

So then you think... what does \(\sqrt{4} = ~? \)

Answer 66

oh

Answer 67

Now I remember :D

Answer 68

: )

Answer 69

Great! So you can think of it either way. See... if you had something like \(\large \sqrt[5]{x^4}\) you could rewrite this as \(\large x^{4/5}\)

Do you see what I mean?

Answer 70

Yes

Answer 71

Cool \(\checkmark\). So what did you get as your limit after pluggign it all in?

Answer 72

let me see

Answer 73

3/4

Answer 74

Awesome! that's correct.

Answer 75

We got 3/5 done, but I got to call it a night. I've got some homework i need to finish up here, so sorry Ican't help youfinish the other 2 :\

Answer 76

No problem,thank you for the help,much appreciated :)

Answer 77

No problem! Good luck :)

Answer 78

You too,take care :)