Question

Need some help with this limit:

Answer 1

\[\lim_{x \rightarrow \infty} \frac{ \ln (x ^{2}+x)-\ln (x ^{2}-3x+2) }{ e ^{\frac{ x+1 }{ x ^{2} }}-e ^{\frac{ 1 }{ x }} }\]

Answer 2

I'm thinking l'hopital's is prolly gonna help.. But we can prolly simplify this too..

Answer 3

I was told to not use L'hopital :(

Answer 4

simplification it is!... ln(A/B)=lnA-lnB

Answer 5

It appears to approach 0.

Answer 6

Uhm let's see.. Idk how far I'll get with this:
let's just focus on the top
\[ln(x^2+x)-ln(x^2-3x+2)=ln(\frac{x(x+1)}{(x-1)(x-2)})\]

Answer 7

The denominator is just e^(1/x^2)) which approaches 1 as x approaches infinity.

Answer 8

The numerator is what bahrom7893 has...and that clearly approaches ln(1) = 0...so you have 0/1 = 0

Answer 9

Actually the bottom approaches 0, no?

Answer 10

no. 1/x^2 approaches 0, but e^0 approaches 1.

Answer 11

\[e^{(x+1)/x^2}\] that is \[e^0\] as x goes to infinity

Answer 12

e^0 - e^0 = 0

Answer 13

\[\lim_{x \rightarrow \infty}\frac{ \ln (\frac{ x ^{2}+x }{ x ^{2}-3x+2 }) }{ e ^{\frac{ 1 }{ x }}(e ^{\frac{ x+1 }{ x ^{2} }-\frac{ 1 }{ x }}-1) }\]

Answer 14

Exacuse me, but the denominator is equivalent to just e^(1/x^2)) which approaches 1 as x approaches infinity.

Answer 15

I don't see why though?

Answer 16

Look at the original problem...that expression in the denominator is equivalent to e^(1/x^2)) by using the law of exponents,

e^x/e^y = e^(x-y) ..use it backwards.

Answer 17

Dude it's a minus

Answer 18

(x+1)/x^2 - 1/x

Answer 19

You can't use the law of exponents there..

Answer 20

I think it would be a good idea to take as common factor e^1/x

Answer 21

Yes, you can..by rewriting it as one exponent!

Answer 22

\[\huge{e ^{\frac{ x+1 }{ x ^{2} }}-e ^{\frac{ 1 }{ x }}}\]IS NOT
\[\huge{e ^{\frac{ x+1 }{ x ^{2} }-\frac{ 1 }{ x }}}\]

Answer 23

oh, no no, I agree but I took it as common factor, not substracting it.

Answer 24

Nah i'm just pointing out why the bottom approaches 0, not 1

Answer 25

hmm taking ln is useless and so is raising to the e..... gah we need to figure out what top approaches..

Answer 26

top is gonna approach 0

Answer 27

the fraction inside ln is going to 1..

Answer 28

Sorry...my error for the denominator.

Answer 29

oh man this is so ideal for l'hopital..... darn it. Np, i thought i was missing something too.

Answer 30

You factored the polynomials inside the Ln?

Answer 31

Can't the first expression in the denominator be facored out by e^(1/x)?

Answer 32

yea i did but it didn't really help, and neither did factoring out e^(1/x).. this isn't even calc at this point, it's just algebra..

Answer 33

Actually factor it out, yea

Answer 34

Isnt it e^(1/x) { e^(1/x^2) + 1}

Answer 35

no it's not. there's a minus in the original

Answer 36

lol too tired to think right now..

Answer 37

So e^(1/x) was factored out with a - instead of our +

Answer 38

I have an idea, I'll work on the denominator:
\[e ^{\frac{ 1 }{ x }}(e ^{\frac{ x+1 }{ x ^{2} }-\frac{ 1 }{ x }}-1)\]
I know that there's a equivalent:
\[e ^{F(x)}-1= f(x)\]
\[f(x)\rightarrow0\]
so it would be:
\[e ^{\frac{ 1 }{ x }}(\frac{ x+1 }{ x ^{2} }-\frac{ 1 }{ x })\]
wich is much easier to operate.

Answer 39

btw the answer is -infinity

Answer 40

I see, so it's -infinity and not 0 ?

Answer 41

Since I made an error in the denominator, I would have to go back to the drawing board. But bahrom7893 says it is - infinity. Although I would like to through that again.