Question

LIMIT sqrt of(2x^2 +1)/(3X+5)
X->infinity

Answer 1

\[\lim_{x \rightarrow \infty} \frac{ \sqrt{2x ^{2}+1} }{ 3x-5 }\]

Answer 2

Use asymptotic behavior, \[\Large 2x^2+1 \sim 2x^2 \\ \Large 3x-5 \sim 3x \]
that would lead to
\[\Large \lim_{x \to + \infty} \frac{\sqrt{2x^2+1}}{3x-5}\sim \lim_{x \to + \infty} \frac{\sqrt{2}x}{3x} \]

Answer 3

Ya, that's because I started by multplying both num and den by the numerator, at the end I got \[3/\sqrt{2}\], and when I checked the answers I found yours, AND I WAS NOW WONDERING Y? THANX , tell me when do we use my method?

Answer 4

Well you can do that, your method is another approach, elimination of the root in the numerator sometimes makes the limit more optical obvious. Asymptotic behavior can always be used when the functions are widely known. In special with polynomials.

You can verify that for yourself:
\[\Large \lim_{x \to + \infty} \frac{2x^2}{2x^2+1}=1 \implies \exists x_0 \in \mathbb{R} : \forall x \geq x_0 , \ 2x^2 \sim 2x^2+1 \]

Answer 5

Or in different words, for some big values of \(x\) there is no longer a difference between \(2x^2\) and \(2x^2+1\). This is what we call asymptotic behavior of functions and we write \(2x^2 \sim 2x^2+1\). Some call this subfield of Calculus "dominance" of functions.

Answer 6

thank you @Spacelimbus

Answer 7

very welcome

Answer 8

but then I don't understand y my method did not work, u know?
ca you please try to get rid of the radical numerator

Answer 9

Your method did not did not work because the limit you're trying to evaluate is not of the required form, you can get rid of roots if you have something like the following:
\[\Large \lim_{x \to + \infty} \sqrt{x^2+x}-x \]because you can change this (without the use of asymptotics, they would not work here) into the form where you use the binomial expression \((a+b)(a-b)=a^2-b^2\)
In your equation the same thing might be possible, but I believe that it will only turn out into a more complicated expression.

However, you can still use Bernoulli de L'Hopital's rule if you want to.

Answer 10

@Spacelimbus thank you once again, WHO CAN KNOW ALL THE MATHS THERE IS IN THE WORLD ha ha ha!

Answer 11

I have yet to meet one who does :-) !