Question

question about l'hospital's rule

Answer 1

\[\lim_{t \rightarrow 0}[t*\ln(1+3/t)\]

Answer 2

this was on my midterm and whats confusing is that it doesn't specifify if it is 0 from the right to the left

Answer 3

\[\lim_{t \rightarrow 0}t \cdot \ln(1+\frac{3}{t}) ?\]

Answer 4

or does it even matter? my friend put limit dne but i had 0 because i thought it turned into an indeterminate product

Answer 5

no it is (1+3)/t

Answer 6

\[\lim_{t \rightarrow 0}t \cdot \ln(\frac{4}{t})?\]

Answer 7

Since 1+3 =4

Answer 8

yeah i guess thats what it turns into

Answer 9

i had that

Answer 10

i jst put it as (3+1)/t cus that was teacher had on exam

Answer 11

That is weird your teacher would it put it like that honestly

Answer 12

I don't mean to sound rude but so it really looks like:
\[\lim_{t \rightarrow 0}t \cdot \ln(\frac{1+3}{t})\] exactly like this?

Answer 13

yup...thats why, it messed me up aswell

Answer 14

no ur not being rude at all don't worry

Answer 15

\[\lim_{t \rightarrow 0} t \cdot \ln(\frac{4}{t})\]
the limit does not exist since from the left t<0 and ln(negative) does not exist

Answer 16

The function is not even defined for t<=0

Answer 17

the limit is zero

Answer 18

Are we doing right limits zarkon?

Answer 19

Is that what we are to just assume?

Answer 20

yay!!

Answer 21

yes...when taking a limit we only look at values of t that are in the domain of the original function.

Answer 22

i figured it out! cus when u graph 4/x both limts from left to right are the same (appraches infinity)

Answer 23

Ok great wise one!

Answer 24

thanks guys

Answer 25

and yeah I agree, my teacher makes troll midterm questions -_-

Answer 26

poorly written question

Answer 27

thanks lol I'll met my teacher know

Answer 28

\[\lim_{t \rightarrow 0}t \cdot [\ln(4)-\ln(t)]=-\lim_{t \rightarrow 0}\frac{\ln(t)}{\frac{1}{t}}\]
\[=-\lim_{t \rightarrow 0}\frac{\frac{1}{t}}{\frac{-1}{t^2}}=\lim_{t \rightarrow 0}\frac{t^2}{t}=\]

Answer 29

just so you know...you can do these problems on wolframAlpha

http://www.wolframalpha.com/input/?i=limit [t*ln%284%2Ft%29]+t-%3E0

Answer 30

\[\lim_{t \rightarrow 0}t=0\]

Answer 31

Assuming t>0

Answer 32

Well and like zarkon says we only look for where the function is defined

Answer 33

that link didn't come out well

just type

limit[t*ln(4/t)] t->0

into wolframalpha

Answer 34

so forget about the assuming part

Answer 35

Zarkon my steps okay?

Answer 36

yes

Answer 37

Thanks King Zarkon! :)

Answer 38

np ;)

Answer 39

Lord Zarkon also works ;)

Answer 40

Ok sorry...Lord Zarkon! I'm bad with how I should address my commanders.
I assume Amideus is out of the woods now?

Answer 41

hmm what if u didn't seperate the logarithm?

Answer 42

It would have looked ugly to me

Answer 43

i got the case "0*infinity" which i just turned into an indeterminate quotient

Answer 44

either way its right right? because i got the same limit.

Answer 45

answer*

Answer 46

and it wasn't that messy...i remember i only had to apply L'hospital's rule once after simplifying

Answer 47

\[\lim_{t \rightarrow 0}\frac{\ln(\frac{4}{t})}{\frac{1}{t}}\lim_{t \rightarrow 0}\frac{\frac{\frac{-4}{t^2}}{\frac{4}{t}}}{\frac{-1}{t^2}}\]

Answer 48

fractions on fractions...now that is messy. :)

Answer 49

\[\lim_{t \rightarrow 0}\frac{\frac{4}{t^2} \cdot \frac{t}{4}}{\frac{1}{t}}\]

Answer 50

\[\lim_{t \rightarrow 0}\frac{\frac{t}{t^2}}{\frac{1}{t}}=\lim_{t \rightarrow 0}\frac{\frac{1}{t}}{\frac{1}{t}}\]
uh oh...

Answer 51

this isn't right...

Answer 52

-1/t^2 on the bottom

Answer 53

on i see... thanks zarkon

Answer 54

or just 1/t^2 since you canceled out the negatives

Answer 55

\[\lim_{t \rightarrow 0}\frac{\frac{4}{t^2} \cdot \frac{t}{4}}{\frac{1}{t^2}}\]

Answer 56

yeah i had t^2 cancel out

Answer 57

\[\lim_{t \rightarrow 0}\frac{\frac{1}{t}}{\frac{1}{t^2}} \cdot \frac{t^2}{t^2}=\lim_{t \rightarrow 0}\frac{t}{1}=\lim_{t \rightarrow 0}t=0\]

Answer 58

than just had \[\lim_{t \rightarrow 0}(4t/4)\]

Answer 59

I like the crazy fraction thing though
it was cute :)

Answer 60

only applied L'H's rule once

Answer 61

but yeh...either way is right huh.. the way i did it was just ineffecient...i'll remember to always simplify logs before differentiating it in all cases for the final :) thanks again guys helped out a lot

Answer 62

that is not true..the limit is zero

Answer 63

for that picture you are correct...that is not the case for this problem though

Answer 64

Limit does not exist at an endpoint!

Answer 65

Unless it is a one-sided limit.

Answer 66

not that those endpoints but it does for thing like \(\sqrt{x}\)

Answer 67

it is really in the definition of a limit

Answer 68

the definition is an if - then statement

Answer 69

more rigorous texts note that x must be in the domain of the original function

Answer 70

Stewart doesn't menmtion it early on in the book , but when you get to the multivariable section he does

Answer 71

see page 923 in the 7th edition

Answer 72

Also ross' book on elementary alanysis also includes it in the definition

Answer 73

also in Marsden/Tromba (vector calculus includes it in the definition)

Answer 74

in stewarts book it says

"If \(f\) is defined on a subset \(D\) of \(\mathbb{R}^n\), then \(\lim_{x\to a}f(x)=L\) means that for every \(\epsilon>0\) there is a corresponding number \(\delta>0\) such that

if \(x\in D\) and \(0<|x-a|<\delta\) then \(|f(x)-L|<\epsilon\)"


it goes on to talk about if n=1 you get the definition of a limit for a single variable function.

Answer 75

again page 923 7th edition (just looking at the stand alone multivariable text...not the full book)

Answer 76

http://www.wolframalpha.com/input/?i=limit&a=*C.limit-_*Calculator.dflt-&f2=t*ln%284%2Ft%29&f=Limit.limitfunction_t*ln%284%2Ft%29&f3=0&f=Limit.limit_0&a=*FVarOpt.1-_**-.***Limit.limitvariable--.**Limit.direction--.**Limit.limitvariable-.*Limit.direction-.*Limit.limitvariable2-.*Limit.limit2-.*Limit.direction2---.*--

Answer 77

First, let me say that I am answering this on the domain of all Reals.
If you are dealing with "all" x which includes the complex numbers then yes your answer would be zero.

"the domain of the function f(x) = √ x is the interval [0, ∞), since negative numbers do not have real square roots and so the limit as x approaches 0 wouldn't exist. However, if the function is explicitly defined for all x (i.e including complex values) then the limit would be 0 as shown in the example below."

http://www.wolframalpha.com/input/?i=limit+x---%3E0+%28x%29^%281%2F2%29

This is the same situation as presented in your problem.

Answer 78

Wolfram Alpha provides a solution on all x...not all real x.

Answer 79

since the domain is \([0,\infty)\) for \(\sqrt{x}\) we only need to consider the positive real numbers...since by definition we only look at the \(x\) values that are in the domain of \(f\). Hence the limit is still zero.

Answer 80

Yes, that I agree with!!!

Answer 81

If the domain is not stated, we assume all real.

Answer 82

if not stated, the domain is usually the larges set of real numbers such that the function is defined.

Answer 83

*largest

Answer 84

I would have to admit that I agree...
If someone were to ask me what is the domain of the square root of x, I would say from 0 to infinity...

I guess, I'll concede.