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lim х²-5х³/х-2х²+6х³
x-∞
lim sin3х/х
х-0
lim (х/х-3)в степени х
х-∞

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вопрос

1) lim (x^2 - 5x^3)/(x - 2x^2 + 6x^3) as x approaches infinity

Dividing all terms by x^3, we get:
lim (1/x - 5)/(1/x^2 - 2 + 6x) as x approaches infinity

As x approaches infinity, the terms with 1/x become negligible, so we are left with:
lim (-5)/(6x) = 0 as x approaches infinity

2) lim sin(3x)/x as x approaches 0

Using L'Hopital's Rule:
lim (3cos(3x))/1 = 3 as x approaches 0

3) lim (x/(x - 3))^x as x approaches infinity

Taking the natural logarithm of the expression first:
lim xln(x/(x - 3)) as x approaches infinity
lim x[ln(x) - ln(x - 3)] as x approaches infinity

Rewriting this as:
lim xln(x) - xln(x - 3) as x approaches infinity

As x approaches infinity, ln(x - 3) becomes negligible in comparison to ln(x), so we are left with:
lim xln(x) as x approaches infinity = infinity * infinity = infinity

Therefore, the limit of (x/(x - 3))^x as x approaches infinity is infinity.