how do you find limit to this question?

[Ahin]

I have been puzzling about this for days:

thanks for help in advance...

[chemistrykind]

geez, this took me a while, but i think the limit is zero. multiply both numerator and denominator by their respective conjugates, and then substitute the 1 once you're done cancelling.













(edited for LaTeX pls dont bully me im not very good yet)

[chemistrykind]

ah, sorry for not adding, but i also find desmos good for times like this!

[1729]

I wouldn't use L'Hopital's rule if you don't need to. Those derivatives and the resulting fractions are nasty to deal with.
Here is the ways you could have approached this problem
 - You could solve through conjugates
 - You could use newton's generalisation and compute the first 3 terms or something
 - You could try logarithms
 - You could use brute force with L'H
 - You can also express each term as taylor i reckon - you will just have to factor out x on some
 - You could use maclaurin series

Edit: Brute force does not work, maclaurin series also doesn't work because it only converges for |x|<1, and taylor doesn't even work because you can't even taylor expand about x=1.

However, I approached it like this:
Limit of f(x)*g(x) = Limit of f(x) * Limit of g(x) if both limits exist and the 1/(sqrt(1+x)-1) exists and the other limit well I assumed to have existed in the first place or if you wanted to be rigourous you could prove a limit existed.
https://bit.ly/3iipMNn - My working out for this approach

Oh that 1-x^2 at the end should be 1+x^2
I should also probably tell you I solved this with the assumption that the limit did exist though, that is alleviated once we actually find the limit of that split up part basically what I'm saying is that you can indeed skip all that long messy stuff, and logs (if you use the log method) and cross multiply. And It's probably easier to directly show that, that limit exists rather than prove boundedness
through epsilon-delta.

Furthermore, I think the conjugate way is easier, its very intuitive to find that the thing splits up and if you are good at algebra it probably is faster like, the 1-x^2 part and then factor it. With conjugates you basically realise that you get the numerator on denominator and denominator on numerator, but also the product of the conjugates which happen to be the same so those cancel out, (A is a product of two expressions and when you differentiate that product (say f and g), you'll get f'g + g'f right
and you know f and g evaluate to 0 at x=1). It simplifies really nicely when you multiply by conjugate of top and the bottom at the same time which is what was originally done in the first place haha

Ngl I only came up with my method b/c wolfram alpha said factor out the denom, when in actuality conjugates are 10 times faster

But if you wanna see how to go with conjugates here - https://bit.ly/2BidcND

But the process as to which you approach this problem is purely subjective.

I hope I helped!! LOL This was quite an interesting problem.

And if you wanna go with multiplying by the other conjugate - https://bit.ly/2ZmNoaU
Just simplify the highlighted and then leave the rest to be the double conjugate multiplication has the effect of removing the zero from the denominator of the sqrt stuff, but then adds some variable stuff and then the variable stuff dies.

[keltingmeith]

I wouldn't use L'Hopital's rule if you don't need to. Those derivatives and the resulting fractions are nasty to deal with.
Here is the ways you could have approached this problem
 - You could solve through conjugates
 - You could use newton's generalisation and compute the first 3 terms or something
 - You could try logarithms
 - You could use brute force with L'H
 - You can also express each term as taylor i reckon - you will just have to factor out x on some
 - You could use maclaurin series

Friendly reminder that L'Hopitals and Taylor expansions are not covered in VCE