Question

HELLLLPPP!
How to solve the following using L-hopital rule?

1) limit h-0 = e^h-1/h
2) limit h-0 = ln(1+h)^1/h

Answer 1

\[\lim_{h\rightarrow0}\frac{e^{h-1}}h\]\[\lim_{h\rightarrow0}\ln(1+h)^{\frac1h}\]

Answer 2

Are these what you mean?

Answer 3

yes

Answer 4

Wait

Answer 5

Did you mean \(\displaystyle\lim_{h\rightarrow0}\frac{e^h-1}h\) for the first one

Answer 6

yes

Answer 7

no no i meant the first one

Answer 8

@kc_kennylau ??

Answer 9

Then it'd be in the form of 1/0

Answer 10

i have to prove the first and second question as equal to 1

Answer 11

Let me ask you again, it's \(\displaystyle\lim_{h\rightarrow0}\frac{e^{h-1}}h\) or \(\displaystyle\lim_{h\rightarrow0}\frac{e^h-1}h\)?

Answer 12

the first one

Answer 13

Then it's unsolvable

Answer 14

You sure it's the first one

Answer 15

yes i am sure it is the limiting case of natural log

Answer 16

@mathmale I can't do an impossible question

Answer 17

L-Hopital's Rule states that if the numerator and denominator both equal zero when the limit is plugged in, then you must take the derivative of the numerator and denominator and then plug in the limit. This will give you the answer. Hope this helps :)

Answer 18

i know but instead of getting 1 i am getting 0.367 :/

Answer 19

Can you attach a copy of your work? Maybe I can see what you may be doing wrong

Answer 20

Forgive me, but I'm going to assume that the problem is actually\[\lim_{h\rightarrow0}\frac{e^{h}-1}h\] If this is the case, then as h approaches zero, e^2 approaches 1 and we end up with 0/0, a perfect case for application of l'Hopital's Rule.

Taking the derivative of numerator and denom. separately,

we get the lim as h approaches 0 of e^h, which is just 1.

@Farheen28: Would you please post a screen shot of the original problem, for verification.

Answer 21

OMG i am so sorry. it is e^h -1/h

Answer 22

@kc_kennylau you were right :P

Answer 23

OK. Problem solved. Note that if the other version were the correct one, l'Hopital's Rule would NOT apply.

Answer 24

If you click the equation button at the bottom of the text box, you can create the equation in a way where there will be no confusion.

Answer 25

@Farheen28 are you able to show a copy of your work?

Answer 26

Well it'd be better to use the "draw" button instead because he/she may not be familiarized with LaTeX.

Answer 27

\[\ln (1+h)^{1/h}\]
this is the other question

Answer 28

Use the formula \(\Large\log_na^b=b\log_na\).

Answer 29

@kc_kennylau the question says to use L'Hopital's rule.

Answer 30

Yes, this is the first step, I'll use the rule later.

Answer 31

Actually, @wotseit, kc_kenny's approach is absolutely correct.

Answer 32

i first used that formula but it gave infinity as the answer
then what should i do?

Answer 33

Oh,whoops, I think I misread what you wrote. Sorry!

Answer 34

No hard feelings

Answer 35

@Farheen28, what do YOU think? Do as much as you can of this work.

Answer 36

Show us what you got after you used this formula @Farheen28

Answer 37

ok wait it'll take some time

Answer 38

\[\frac{ 1 }{ h }\ln (1+h)\]
\[\frac{ 1 }{ h }.\frac{ 1 }{ 1+h }.1\]
then i put the limit and got infinity as my answer

Answer 39

\(\dfrac1h\ln(1+h)\) is in the form of ∞×0...

Answer 40

You can't apply l'hopital rule to ∞×0

Answer 41

It seems that instead of taking the derivative of h (which would be the denominator), you left it as 1/h and multiplied that by the derivative of ln(1+h). Don't forget to take the derivative of both the numerator and the denominator when applying L'Hopital's Rule

Answer 42

are these two different \[\infty \times0\] and \[0\times \infty\]
no right?
because you can apply l hopital to \[0 \times \infty\]

Answer 43

And no you can't apply l'hopital to \(0\times\infty\)

Answer 44

You need to convert \(0\times\infty\) into \(\dfrac00\) before you can use it.

Answer 45

Where is infinity coming from? Just curious.

Answer 46

\(\dfrac1h\) itself is infinity

Answer 47

i am confused

Answer 48

You can't apply l'hopital to \(\dfrac1h\ln(1+h)\), but you can definitely do so to \(\dfrac{\ln(1+h)}h\).

Answer 49

Although they are basically the same

Answer 50

I don't know anything about infinities but if we multiply 1/h by ln (1+h) we get
\[\frac{ \ln(1+h) }{ h }\]
And then you just apply L'Hopital's rule
And kenny beat me again. I'm so slow haha

Answer 51

could you elaborate how am i going t

Answer 52

to reach 1 as my answer?

Answer 53

Apply L'Hopital's rule to
\[\frac{ \ln(1+h) }{ h }\]
and you will get 1. I can walk you through it if you'd like.

Answer 54

after differentiating it i will get \[\frac{ 1 }{ 1+h }\] right?

Answer 55

@wotseit ?

Answer 56

Yes, you are right. Keep going!

Answer 57

then when i will put 0 in h i will get 1 as my answer :)

Answer 58

Exactly :D

Answer 59

thank you all :D
one more thing can i apply l hopital rule when \[0^{0}\], \[0 \times \infty\] and \[1^{\infty}\] ?

Answer 60

No, l'hopital rule only works for \(\dfrac00\) and \(\dfrac\infty\infty\)

Answer 61

However, you can convert the above cases into \(\dfrac00\) and then you can use it

Answer 62

how can i do that?

Answer 63

\[\Large0^0=e^{\ln0^0}=e^{0\ln0}=e^{\frac{\ln0}{\infty}}=e^{-\frac\infty\infty}\]

Answer 64

Tell me which step you don't understand if there's any

Answer 65

3rd

Answer 66

It's using the formula \(\log a^b=b\log a\)

Answer 67

If that's what you mean by the "3rd" step

Answer 68

4th actually

Answer 69

Because \(\dfrac1\infty=0\)

Answer 70

\[\Large0^0=e^{\ln0^0}=e^{0\ln0}=e^{\frac1\infty\ln0}=e^{\frac{\ln0}{\infty}}=e^{-\frac\infty\infty}\]

Answer 71

Think of it like this: infinity is a very large number. If you divide a number by a very large number, you get very close to zero. That's how I like to think of it, at least.

Answer 72

ah now i get it :)
i am sorry i put you in so much trouble :/

Answer 73

no problem :)

Answer 74

\[\Large1^\infty=e^{\ln1^\infty}=e^{\infty\ln1}=e^{\infty0}\]

Answer 75

\[\infty\times0=\frac10\times0=\frac00\]

Answer 76

thank you :D

Answer 77

No problem :)

Answer 78

And as always, \(\infty\times0=\infty\times\dfrac1\infty=\dfrac\infty\infty\)