Question

\[\LARGE \lim_{x \rightarrow \infty} (\frac{x^2+5x+3}{x^2+x+3})^x\]

Answer 1

is it ^x or * x

Answer 2

^x

Answer 3

Only for numerator?

Answer 4

no..whole

Answer 5

Can we apply L-Hospital's Rule..

Answer 6

@DLS

Answer 7

yes

Answer 8

Then...Differentiate Numerator and denominator Seperately..)

Answer 9

e^4
e^2
e^3
1

Answer 10

@Yahoo! didnt get the ans

Answer 11

Use this
\[ \huge a = e^{\log a}\]
then use L'Hopital

Answer 12

:O
what in the world is that??

Answer 13

Just use your logic. We know that \(\large lim_{x \rightarrow \infty}\dfrac{x^2}{x^2} = 1\)

So \(\large \lim_{x \rightarrow \infty}\left(\dfrac{x^2+5x+3}{x^2+x+3}\right)^x = \lim_{x \rightarrow \infty}1^x = \boxed{1}\)

Answer 14

logic fails
answer isnt 1

Answer 15

And how do you know?

Answer 16

@geerky42 ^x is only for numerator

Answer 17

its for whole of it

Answer 18

not only num

Answer 19

@geerky42 relax..answer isn't 1..

Answer 20

Well, how do you know?

Answer 21

Lol....u said "no..whole" hm..hm

Answer 22

lol i said no..not num..whole

Answer 23

oh...then...i was mistaken..)

Answer 24

@Yahoo! try that

Answer 25

And what in tarnation kind of class are you taking, anyway?

Answer 26

but my answer too ended up in 1..@experimentX

Answer 27

How do you know answer isn't one? @DLS

Answer 28

i have the solution? :/

Answer 29

Why do you need our help if you have solution?

Answer 30

I DIDNT UNDERSTAND IT ?

Answer 31

http://www.wolframalpha.com/input/?i=lim+x-%3Einf+x+log%28%28x^2%2B5x%2B3%29%2F%28x^2%2Bx%2B3%29%29
e^4

Answer 32

correct

Answer 33

@geerky42 hope that satisfied you.

Answer 34

You mean answer key, not solution...

Answer 35

Do you have to show your work?

Answer 36

they have done weird things
\[\LARGE \lim_{x \rightarrow \infty} [(1+\frac{4x+1}{x^2+x+2})^{\frac{x^2+x+2}{4x+1}}]^\frac{(4x+1)x}{x^2+x+2}\]

Answer 37

What in tarnation kind of class are you taking?

Answer 38

Since it appears that we cannot solve it, just use WolframAlpha. Good luck...

Answer 39

Don't use "We".

Answer 40

this is straight forward L'Hopital rule
put that base in log an raise it to e ..
then you get type inf x 0 ... use L'hopital rule couple of times, you should get answer

Answer 41

still ends up complicated :/

Answer 42

how to do you evaluate
\[ \lim_{x \to \infty} x \log\left( x^ 2 +5x+3 \over x^2 + x+ 3\right)\]

Answer 43

wait i guess i got it :O

Answer 44

thanks! :D

Answer 45

for all functions of type
\[ \huge f(x)^{g(x)} = e^{\log f(x)^{g(x)}} = e^{g(x) \log f(x)}\]