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

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

\[\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\]
I assume that i would have to multiply both numerator and denominator with a conjugate
correct

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

and that conjugate would be \(\large\sqrt{9+h}+3\)
alright. At least that assumption was correct
so from there, since i would multiply \(\sqrt{9+h}-3\times \sqrt{9+h}+3\)
since \(\sqrt{x^{2}}\) is equal to \(x\), then \(\sqrt{9+h}\times \sqrt{9+h}\) would be equal to \(9+h\)
Since the middle terms cancel out, I believe we are left with \(\Large\lim_{h \rightarrow 0} \frac{h}{h(\sqrt{9+h} +3}\)
in the numerator 9+h-9 in the denominator ? you are correct.
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
\(\Large\lim_{h \rightarrow 0} \frac{\color{red}h}{\color{red}h(\sqrt{9+h} +3}\) The things in red will cancel out, so you should be left with \(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\)
@nincompoop from here, would I just plug in zero in place of h? (assuming, of course, that I hadn't taken any erroneous steps)
I didnt read the entire steps above but if the denominator doesnt go to 0 you can plug in 0 for h.
at this point, it doesn't.
took a lot of simplifying, which I pray I got correct XD. Thanks for the tip, movie buddy :3
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
you need to show me the original function tho
alright. I've already verified that most of my steps were correct with surji, and here is the original fxn \(\Large\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right)\)
@AravindG I think I already outlined that somewhere up there :P
You got this :) You just need to stay confident!
thanks :) @Jhannybean
Yeah you are right. The limit exists.
\[\boxed{\lim_{h \rightarrow 0}\left( \frac{ \sqrt{9+h}-3 }{ h } \right) = \frac{1}{6}}\]
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
thanks, jhan
I knew there was something odd LAUGHING OUT LOUD
jhan :*
Alright. Good to know. \(\Large\lim_{h \rightarrow 0} \frac{1}{\sqrt{9+h} +3}\) \(\Large\frac{1}{\sqrt{9+0} +3} = \frac{1}{3+3}\)
Yep.
and that would give you \(\Large\frac{1}{6}\). Alright then, that makes sense!
So at that point, you can plug in 0, since it wouldn't become undefined. g'nice
ye why is that?
why can't we just plug it in immediately?
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
interesting
At least I think that's what the prof. said o-o lol
so what does it mean that the limit = 1/6 ?
why dost thou question me profusely? XD
fundamental concepts and analysis
I see o-o
That means at some value of x, the function will approach y at 1/6,
^that's the technical meaning of a limit
and then what?
At this point, I'm not sure what you're asking
haha
Seriously XD I can't keep track of all these questions. What more do you want than the technical definition? o-o
just telling me what the formal definition of limit does not tell me what you understand about it, that's all
Something to do with reimann sum right? :P lol
Anyhow, thank you, I'll close this for now :)
@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
they are related
I see

Not the answer you are looking for?

Search for more explanations.

Ask your own question