how do you take the limit of a radical function if x is under

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673B, 86x46px

wrong one whoops

What the x value you're asked to find the limit of?

>>8385515
> take the limit

Are you asking how to take the derivative or just the limit, if the limit then what limit are you trying to evaluate?

>>8385530
the limit as h approaches infinity.

>>8385518
i dunno, i see the top is just equal to x + 2 and then you can separate the two terms into x/(bottom) + 2/(bottom)

maybe that will simplify things

>>8385534
>top is just equal to x + 2

No it isn't.

>>8385532
So, lim h-> infinity of [f(x+h)-f(x)]/h?
That would be the derivative of f(x)


I'm assuming based on your question you're in calc 1, so I'll try to be as laymen as possible.

Here you can apply the quotient rule by defining the top of
>>8385518
as "f(x)", and the bottom as "g(x)" so this would simply return (f’(x) g(x) − g’(x) f(x) ) /(g^2)

Which simplifies to:

(x)/(sqrt(x^2 + 4) (sqrt(3x+7) + 9)) - (3 sqrt(x^2 + 4)) / (2 sqrt (3x+7) (sqrt(3x + 7) + 9) ^2)

I don't know how to use the math plugin because i exclusivly lurk.

If you're looking for the expansion and cancellation of all those h's, then hopefully someone else will post it because I'm WAYYY too lazy to do all that work.

>>8385557
what if you also had to show the limit as x itself goes to infinity?

>>8385563
Showing it on paper is kind of weird but just look at the highest x term in the equation. sqrt(x^2) > sqrt(3x) your ass is going to infinity at infinity.

>>8385563
If x is going to infinity then that is an end behavior basically.

To put it simply (math major's pls don't kill me I'm engi) because infinity is fucking big and sqrt(infinity) is infinity for all intents and purposes, because infinity is so large we can also neglect the other constants we have.

so we basically have infinity^2 over infinity, which is basically infinity.

So as x - > infinity f(x) goes to infinity

>>8385567
but from what i was told by my prof infinity/infinity is of indeterminate form and so you have to say there's no limit or you have to find a way to cancel it out by factoring out x or h.

like in the case of x+1/x+1 you can say it's infinity/ over infinity but it's actually 1/1 because you cancel it out and just say x's domain is such that it can't have -1. or am i wrong?

>>8385569
It's indeterminant because it's a bitch to cancel the radicals but common sense tells you the limit of sqrt(x^2+4)/sqrt(3x+7) is approaching infinity at infinity.

>>8385569
If you divide everything by x, the leading term, you should be able to cancel out some terms on the bottom. The terms left with a constant over x are going to be basically zero because x is going to infinity. Doing this you're left with the answer being infinity

However only do this if the limit is going to infinity

>>8385569
You have infinity^2 over infinity for your radical function in the OP.

Or
Lim x - > infinity (x^2 / x) which is:
lim x - > infinity (x/1)
lim x - > infinity x
infinity

>like in the case of x+1/x+1
lim x - > infinity (x+1/x+1) is infinity/infinity which is an indeterminate form.


Without finding the oblique asymptotic we can actually redefine the function in an alternate form of (x^2 + x + 1) / x which means when we take the limit of x tending to infinity we get f(x) going to infinity and x tending to negative infinity f(x) tending to negative infinity.

>>8385582
WAIT I'M DUMB

It's not x+1 + 1/x

So (x+1)/(x+1) would be a point on a number line at 1. Which is why it has no end behaviors, which is why it returns indefinite forms for x tending to either infinity.

>>8385515
I think our teacher gave you the solution.

If you're still in school

>>8385557
>lim h-> infinity of [f(x+h)-f(x)]/h is the derivative
are you actually retarded or just pretending

>>8385518
well you didn't say what x value you're approaching, so i'll assume you mean as x approaches x. in that case, it is simply f(x).

>>8386462
no it's x approaches infinity. my problem is that infinity/infinity as my teach said was "useless from an analytical standpoint"; it tells us nothing so we need a way to cancel it out. so he showed me how i can just multiply by 1 via sqrt/sqrt of the above term and it lets out the x.

but my problem is, how do i get the denominator to rationalize? it's still stuck...

>/sci/ unironically helps people with homework now

4chan sure has changed for the worse

>>8387600
>helping by not being able to find the limit

lmao

If you know what the graph of 1/x looks like, you can derive the value that way. It's not easy to solve this limit algebraically

>>8386327
Lmao I didn't even notice that. It was like 4am.

I'll take pretending for 400, Alex.

wth i keep getting the wrong answer. the correct answer is 1/sqrt(3) according to wolfram alpha by l'hopital's. but i assume you can't use it yet.

Multiply by the conjugate:
[math] \sqrt{3x+7} - 9 [\math]
Then you get something with x^3 on the top, which is the highest term. The highest term on the bottom is an x^2 term. so the answer i get is infinity by looking at the degrees of the x terms.

>>8385515
You can rip the x out of the radicals by deviding the inside by x^2 or something and putting the x on the outside. Do it on the bottom too so can cancel them out. Then you sould have something like sqrt(1+1/x) which would just be sqrt (1) as x->infinity