Question

please explain!!!

Answer 1

\[\lim_{h \rightarrow 0} (e^h-1)/h\]

Answer 2

What do you want us to explain?

Answer 3

how to solve the limit.

Answer 4

Okay so first let's just plug in 0. What do we get?

Answer 5

indeterminate
form

Answer 6

0/0

Answer 7

That's an indeterminate form.

Answer 8

What rule can we use to help us solve indeterminate forms? L'Hospital's rule right?

Answer 9

my guess is that you not know l'hopital yet, and are just beginning calculus
this limit is \(1\) and the reason is, one definition of \(e\) is that it is the number that makes this limit 1

Answer 10

@Fgcbear16 Do you know l'Hopital's rule?

Answer 11

yea i don't know l'hopital yet

Answer 12

not if you don't know it yet. this is an introductory problem
if you know how to take a derivative, then this is defined to be the derivative of \(f(x)=e^x\) at \(x=0\)
if you know that, and you know that the derivative of \(e^x\) is in fact just \(e^x\) then you get \(e^0=1\)
but if you do not know that, then the only thing you can do is try a numerical exercise
that is plug in numbers closer and closer to 0 and see that your answer is closer and closer to 1

Answer 13

I am having connection problems just so you know.

Answer 14

how'd i guess? then there is not much you can do, other than check with numbers like \(x=.01,x=.01, x=.001\) and see what you get

Answer 15

yeah connection is bad

Answer 16

Have you learned derivatives? Have you only learned limits?

Answer 17

I do know derive.

Answer 18

The problem was to prove the derivative of e^x

Answer 19

\[e^x[\lim_{h \rightarrow 0}(e^h−1)/h ]\]

Answer 20

That how far I got

Answer 21

Oh, then maybe you need to go about it differently?

Answer 22

Oh, what you could do is consider that @dpaInc did before.

Answer 23

i thought of that but really didn't want to plug in numbers. i prefer algebra or concepts.

Answer 24

i know i am lazy

Answer 25

Have you learned squeeze theorem?

Answer 26

yes

Answer 27

Can you think of any functions which will squeeze this guy into 0?

Answer 28

\[2^h \le e^h \le 3^h\]

Answer 29

But you also need to be sure you know the limit of \(2^h/h\)

Answer 30

And remember it is \(e^h-1\)

Answer 31

i know

Answer 32

There is also the delta epsilon method.

Answer 33

\[ 2^h-1≤e^h-1≤3^h-1\]

Answer 34

\[(2^h−1)/h≤(e^h−1)/h≤(3^h−1)/h\]

Answer 35

But how are those limits any easier?

Answer 36

there are not i just tried it. life stinks

Answer 37

i guess i will just plug it in. thanks for the help!

Answer 38

No problem.

Answer 39

If you knew l'hopital's rule this would have been very easy.

Answer 40

i wish i did. oh well ttyl