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or you can solve it using e and ln too

wait....that's not the case here, what's the function?

Could you please explain how to come to that step?

wrong, my last post is not true

I just put exp(ln(.)) around the function and used logarithm rules.

That alright?

Ok, al right

What is it you don't understand?

Its ok, I understand what you said

I have something else to ask

I just wish to ask, what are the things to consider while solving problems on limit and continuity

I'm sorry, but I don't know a general answer to that question. Analysis is a huge field.

No no, you don't have to go to that. I am in high school

iam, I can respond now if you're around.

Yes I am online

Then\[\lim_{x->\infty}\log y=\lim_{x->\infty}\frac{\log \left( \frac{f(x+1)}{f(1)} \right)}{x}\]

Indeterminate, 0/0 (change that limit to 0, not infinity).

No need to reply to this message. Just informing that I have received the notes

\[=\frac{f'(1)}{f(1)}=\frac{6}{3}=2\]

I have got that. I can handle the rest.

That is\[\lim_{x \rightarrow 0}\log y = 2 \rightarrow \lim_{x \rightarrow 0}y=e^2\]

Okay

You are really a good person. In this world no one does so much for any other without any return

Sometimes they do.

Ok let me study the whole thing,
I will let you know when I need any help.
Thank you sooooooooo much

Is that your university?

No.

Anyhow thanks

np. good luck

I don't want to be nosy about you, as it may offend you

After all I am receiving something GREAT from you

It's okay. Just do well in STEP.

It will be a gift from me to you

I will do good, and thats a promise

Well, it's worth it then :)

Ok bye now

See you.