Solve consult limit n·(e^-n)

The value of the limit is 0.

Step-by-step explanation:

We are given the following limit:

Simplifying the function:

Replacing infinity at the numerator and denominator, we get infinity divided by infinity. Thus, L'Hospital Theorem can be applied, which means that we find the limit of the derivative of the numerator and the denominator.

Derivative of n is 1, of is

Then

The value of the limit is 0.

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."

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