Calculus Help! lim x-->infinity (x+3)/((√2x^2-7))

= (√2)/2. thats the answer, I just need to know how to do the steps.


thanks in advance!!

(x + 3) / (√(2x^2 - 7))

As x -> ∞, your equation will start to look like this:
(Very large number + 3) / (Very large number - 7)

Because 3 and 7 are insignificant compared to these very large numbers, it's okay to drop them out of the equation entirely, so we come up with:
x / √(2x^2)
x / (x * √2)

The x cancels and we get:
1 / √2

If you multiply by √2 / √2 (which equals 1, so you're not changing the value of the fraction):
√2 / (√2 * √2)
√2 / 2

Possible intermediate steps:

lim_(x->infinity) (x+3)/(2x-7)

Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have,

lim_(x->infinity) ((d/dx (3+x)))/((d/dx (-7+2x))) = lim_(x->infinity) 1/2 = 1/2

lim x-->infinity (x+3)/((√2x^2-7))
= lim x-->infinity (x)/((√2x^2)), after neglecting lower degree terms
= 1/√2
= (√2)/2

Well, 1+1 = 2

So start from there.