The value of limx→∞ x3sin(1/x)−2x21+3x2 is __________

To solve this limit, we can use L'Hopital's Rule, which states that the limit as x approaches a of [f(x) - g(x)] / [h(x) - k(x)] is equal to the limit as x approaches a of [f'(x) - g'(x)] / [h'(x) - k'(x)], provided the latter limit exists.

First, let's rewrite the expression as a fraction:

lim(x→∞) [x^3sin(1/x) - 2x^2] / [1 + 3x^2]

Now, we can differentiate the numerator and the denominator:

Numerator derivative: 3x^2sin(1/x) - x^3cos(1/x)/x^2 - 4x Denominator derivative: 6x

Applying L'Hopital's Rule, we get:

lim(x→∞) [3x^2sin(1/x) - x^3cos(1/x)/x^2 - 4x] / 6x

As x approaches infinity, sin(1/x) approaches 0 and cos(1/x) approaches 1. Therefore, the limit simplifies to:

lim(x→∞) [3x^20 - x^31/x^2 - 4x] / 6x = lim(x→∞) [-x - 4x] / 6x = -5/6

So, the value of the limit is -5/6.