URGEEEEENT HELP!! Determining the limit :(

Determine the limit

1) lim x^2 +3x -2 / 3x^2 +4x - 1
when x ---> infinity

The answer will be 1/3

2) lim -x/ sqrt 4+ x^2
when x ---> infinity

The asnwer will be 1

Thank you!!!

lim x→∞ (x² + 3x - 2)/(3x² + 4x - 1) = ∞/∞
Either apply L'Hopital's rule or divide all terms by the highest power of x. If you don't know the first option, here's how to do the second:
lim x→∞ (1 + 3/x - 2/x²)/(3 + 4/x - 1/x²) = (1 + 0 - 0)/(3 + 0 - 0) = 1/3

L'Hopital's:
lim x→∞ (2x + 3)/(6x + 4) = ∞/∞
lim x→∞ 2/6 = 1/3

--------------------------------------
lim x→∞ (-x)/(√(4 + x²)) = -∞/∞

Take a look at the denominator first:
√(4 + x²)
√(x²( 4/x² + 1 ))
√x² ∙ √(4/x² + 1)
x √(4/x² + 1)

lim x→∞ -x/(x√(4/x² + 1))
lim x→∞ -1/√(4/x² + 1) = -1/√(0 + 1) = -1

Your answer to #1 is correct.

I found the answer to #2 to be -1, not 1.

lim_(x->infinity) -x/sqrt(x^2+4)

Factor out constants:

= -(lim_(x->infinity) x/sqrt(x^2+4))

Simplify radicals, x/sqrt(x^2+4) = sqrt(x^2/(x^2+4)):

= -(lim_(x->infinity) sqrt(x^2/(x^2+4)))

Using the power law, write lim_(x->infinity) sqrt(x^2/(x^2+4)) as sqrt(lim_(x->infinity) x^2/(x^2+4)):

= -sqrt(lim_(x->infinity) x^2/(x^2+4))

Indeterminate form of type infinity/infinity. Using L'Hospital's rule we have, lim_(x->infinity) x^2/(x^2+4) = lim_(x->infinity) (( dx^2)/( dx))/(( d(4+x^2))/( dx)):

= -1