Question

Check my work please. Not sure if I'm going in the right direction here

Answer 1

\[\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\]

Answer 2

I assume that i would have to multiply both numerator and denominator with a conjugate

Answer 3

correct

Answer 4

and that conjugate would be \(\large\sqrt{9+h}+3\)

Answer 5

alright. At least that assumption was correct

Answer 6

so from there, since i would multiply \(\sqrt{9+h}-3\times \sqrt{9+h}+3\)

Answer 7

since \(\sqrt{x^{2}}\) is equal to \(x\), then \(\sqrt{9+h}\times \sqrt{9+h}\) would be equal to \(9+h\)

Answer 8

Since the middle terms cancel out, I believe we are left with

\(\Large\lim_{h \rightarrow 0} \frac{h}{h(\sqrt{9+h} +3}\)

Answer 9

in the numerator 9+h-9
in the denominator ?
you are correct.

Answer 10

Nice! So I wasn't sure if I was multiplying things correctly

oh, and I kind of forgot to close the parentheses in the denominator, but other that, it looks good to you? @surjithayer

Answer 11

\(\Large\lim_{h \rightarrow 0} \frac{\color{red}h}{\color{red}h(\sqrt{9+h} +3}\)

The things in red will cancel out, so you should be left with

\(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\)

Answer 12

@nincompoop from here, would I just plug in zero in place of h? (assuming, of course, that I hadn't taken any erroneous steps)

Answer 13

I didnt read the entire steps above but if the denominator doesnt go to 0 you can plug in 0 for h.

Answer 14

at this point, it doesn't.

Answer 15

took a lot of simplifying, which I pray I got correct XD. Thanks for the tip, movie buddy :3

Answer 16

assuming that you've done correctly up to the last portion, all you have to do is plug in the value of h = 0 and simplify

Answer 17

you need to show me the original function tho

Answer 18

alright. I've already verified that most of my steps were correct with surji,

and here is the original fxn
\(\Large\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\)

Answer 19

@AravindG I think I already outlined that somewhere up there :P

Answer 20

You got this :) You just need to stay confident!

Answer 21

thanks :) @Jhannybean

Answer 22

Yeah you are right. The limit exists.

Answer 23

\[\boxed{\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right) = \frac{1}{6}}\]

Answer 24

there are few key points that you probably already learned whether or not limit may or may not exist based on the function itself without doing all of the work.
it cuts many of your work

Answer 25

thanks, jhan

Answer 26

I knew there was something odd LAUGHING OUT LOUD

Answer 27

jhan :*

Answer 28

Alright. Good to know.

\(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\)

\(\Large\frac{1}{\sqrt{9+0} +3} = \frac{1}{3+3}\)

Answer 29

Yep.

Answer 30

and that would give you \(\Large\frac{1}{6}\). Alright then, that makes sense!

Answer 31

So at that point, you can plug in 0, since it wouldn't become undefined. g'nice

Answer 32

ye why is that?

Answer 33

why can't we just plug it in immediately?

Answer 34

because then the whole fraction would be set over 0. As long as the denominator is zero, we conclude that the limit does not exist, when in reality it does

Answer 35

interesting

Answer 36

At least I think that's what the prof. said o-o lol

Answer 37

so what does it mean that the limit = 1/6 ?

Answer 38

why dost thou question me profusely? XD

Answer 39

fundamental concepts and analysis

Answer 40

I see o-o

Answer 41

That means at some value of x, the function will approach y at 1/6,

Answer 42

^that's the technical meaning of a limit

Answer 43

and then what?

Answer 44

At this point, I'm not sure what you're asking

Answer 45

haha

Answer 46

Seriously XD I can't keep track of all these questions. What more do you want than the technical definition? o-o

Answer 47

just telling me what the formal definition of limit does not tell me what you understand about it, that's all

Answer 48

Something to do with reimann sum right? :P lol

Answer 49

Anyhow, thank you, I'll close this for now :)

Answer 50

@nincompoop

Is the reimann sum the same as the epsilon-delta limit definition?! Because if it is, then we barely went over that in class. And if they are indeed the same thing, then I do understand what you're saying

Answer 51

they are related

Answer 52

I see