Question

@lovekblue \[\lim_{x\rightarrow \infty} \frac{2^{x/2}}{3^{x/3}}\]

Answer 1

notice how l'hop will fail?

Answer 2

yes, cuz i've tried it LOL

Answer 3

since it is a 2^x , u will always get 2^x back

Answer 4

2^(x/3)*

Answer 5

yep.

Answer 6

problems of this type, one-term-numerator and one-term denominator, combine them as a fraction raised to a common power.

Answer 7

will remember that!

Answer 8

can u show me some (inf)^ inf

Answer 9

inf^inf = inf

Answer 10

I have one here, but it looks pretty mesy

Answer 11

#12?

Answer 12

yes

Answer 13

that's 0^inf

Answer 14

after applying ln,
ln y = lim x ln (3x/1+2x2])=(inf)(-inf)=-inf
y = e^-inf = 1/e^inf = 1/inf = 0
no L'hop needed.

Answer 15

\[\Large r^\infty =\begin{cases}0, & 0<r<1\\ \infty, & r>1\end{cases}\]

Answer 16

(inf)(-inf)=-inf < where did u get this from?

Answer 17

lim (3x) / (1+2x^2) = 0

lim ln [ (3x) / (1+2x^2) ] = -inf

Answer 18

i still dont get how is 3x / 1 +2x^2 = 0
dont u sub inf in x?

Answer 19

or is it because the degree of x is bigger in denominator?

Answer 20

ln (inf) = inf
ln (0) = -inf
e^inf = inf
e^-inf = 0

Answer 21

BINGO!!! it's all about the degree.

Answer 22

omg yes, i finally got something xD

Answer 23

it's been a while since u last said bingo LOL

Answer 24

a lot of students go through L'Hops evaluating lim (x-->inf) f(x)/g(x) for algebraic functions f and g, without realizing that only the degree of f and g determine whether the answer is inf, some constant c, or 0.

Answer 25

consider lim (x->inf) [(7x^7 + 6x^6 - 1) / (9x^9-5)]

Answer 26

it would be a waste applying L'hop 7 times and conclude that lim = 0

Answer 27

oh 0 !

Answer 28

yea, my prof never taught us that either

Answer 29

just write it as (0x^9 + 7x^7 + 6x^6 - 1) / (9x^9 - 5)
the answer is determined completely by the coefficients of the terms with the highest degree
0/9 = 0

Answer 30

oh, so if it is 5 x^9 .... / 6 x^9 , then limit is 5/6

Answer 31

YES!

Answer 32

do u think i'm ready for the exam >.> 26 more hrs ..

Answer 33

you need to double-check your solution, since you make mistakes plenty of times

Answer 34

yea.. I should get at least 10% higher than what i have if i didnt make mistakes :/

Answer 35

and for L'H, so do we first look for 0^ inf = 0 or inf ^ 0 = 1 first. If it is not there, then try to manipulate the function into inf/inf or 0/0?

Answer 36

right. L'Hop works only on inf/inf or 0/0

Answer 37

and if the exponent is not 0 or inf, use ln to bring it down.

Answer 38

right. but make sure to write an equation before using ln.

Answer 39

why..? i always ignore the y @@

Answer 40

yes.

Answer 41

For lim x-> 0+ , e^ 1/x = inf ?

Answer 42

isnt 1/0 u.d?

Answer 43

lim (3x/1+2x2)^x = 0^inf. write the equation first.

y=lim (3x/1+2x2)^x

then take the ln of both sides

Answer 44

or is it because 0+ or 0-
0+ is inf
0- is -inf

Answer 45

1/0 is undefined. in limit problems, 1/0 = inf

Answer 46

why do u still need ln it?
dont we know that if we have 0^inf form, it must be 0 for sure.

Answer 47

ooo

Answer 48

better safe than sorry.

Answer 49

haha true, especially it is me who often make stupid mistakes

Answer 50

last midterm, the ques asks for vertical asym, and i circled the one talking abt hori asym..>.>

Answer 51

it's like i can never change this "careless" personality :(

Answer 52

\[\frac{ +\infty }{ 0 }\] <what will this form look like?

Answer 53

My note says "Very small # approaches -inf" , see if it approach from left or right"

Answer 54

inf (1/0) = inf (inf) = inf

Answer 55

oh ok ,thanks. u are just too awesome :P

Answer 56

I'm done my ques! will u be online anytime in the uncoming 26hrs to take last min ques @_@

Answer 57

i'll try.

Answer 58

ok thanks!! but I think I just need to practice more and should be good
thanks a lot alot :DDDDD

Answer 59

Give all my hard ques to ur students and fail them ^^

Answer 60

i will. hahaha.

Answer 61

rest time...

Answer 62

im gonna go coffee time :)

Answer 63

see u later then :D

Answer 64

@sirm3d you said inf / 0 = inf

But in \[\frac{ e^x }{ \ln(1-\frac{ 2 }{ x } )} . \]

The limit is - inf not +ve inf, why?

Answer 65

How do I know which case in inf / 0 is +ve or -ve?

Answer 66

you only need to determine if it's 0+ or 0-, then follow the rule on signs.
(-inf)/(0+)=(-/+) inf/0=(-) inf

Answer 67

how do i know if it is 0+ or 0- ?

Answer 68

Here i think is 0+ ?