Limx - > 2 [x ^ 3 + ax + B / X-2] = 8 for ab

Limx - > 2 [x ^ 3 + ax + B / X-2] = 8 for ab

Let limx → - 1 (x ^ 3 + ax ^ 2-x + 4) / x + 1 = B (B is a non-zero constant), find a, B

lim(x→-1 )(x^3+ax^2-x+4)/(x+1)=b
So x = - 1 molecule x ^ 3 + ax ^ 2-x + 4 = 0
arcsinx-x
-1+a+1+4=0
a=-4
lim(x→-1 )(x^3+ax^2-x+4)/(x+1) (0/0)
=lim(x→-1 ) 3x^2+2ax-1
=3-2a-1
=10=b

Finding limit limx → 1 x / (x-1)

1

Limx ^ (2) e ^ (1 / x ^ 2) x tends to 0, the law of lobita

limx^2*e^(1/x^2) = lime^(1/x^2)/(1/x^2)
= lime^(1/x^2)(-2/x^3)/(-2/x^3)
= lime^(1/x^2) = + ∞.

Finding limit LIM (x → 0) (sin 4x) / (SIN) with lobita's rule

Limx tends to infinity ln (1 + 1 / x) / arccot x limit by lobita's law

The limit limx → positive 3ln (x-3) / ln (e ^ x-e ^ 3) is calculated by the lobida rule

limx->3 1/(x-3)/[e^x/(e^x-e^3)]=limx->3 [e^x-e^3]/e^x(x-3)=limx->3 e^x/[e^x(x-3)+e^x]=e^3/e^3=1

Limx tends to infinity x ^ 2 (1-cox1 / x) = what

Limx x →∞ (x ^ 2) - limx x →∞ (x ^ 2) =? Is infinity minus infinity equal to zero

Your subtraction is the same as the subtracted. Of course, it's 0
But if there is something in the title that you can calculate, it is definitely not zero. The general practice is to multiply by a fraction whose numerator denominator is the sum of the subtracted and the subtracted, and then use the square difference formula for the denominator. If not, use the lobita rule

Find limx → 0 ln (1 + x) / X
Please give a detailed idea and process, it's better to give a good idea and method to solve the function limit, thank you

x-->0 ln(1+x)-->x
lim(x-->0)ln(1+x)/x
=lim(x-->0)x/x
=1