Calculus 1 Evaluating limit help please :(

lim ( x^3 - x)/√ (3x+1) - 2x
x--->1

P.S: -2x is not included under square rooot
thank youuu so mucch for your time

As you wrote it -2x isn't in the denominator either, but the answer would be -2 and just involves plugging in, so I don't think that's what you meant.

lim (x^3 - x) / (sqrt(3x + 1) - 2x)
x--> 1

lim (x(x^2 - 1)) (sqrt(3x + 1) - 2x)
x --> 1

lim (x(x + 1)(x - 1)) / (sqrt(3x + 1) - 2x)
x --> 1

Multiply top and bottom by sqrt(3x + 1) + 2x

lim (x(x + 1)(x - 1)(sqrt(3x + 1) + 2x) / (3x + 1 - 4x^2)
x--> 1

lim(x(x + 1)(x - 1) (sqrt(3x + 1) + 2x)) / ((-4x - 1)(x - 1))
x--> 1

lim(x(x +1)(sqrt(3x + 1) + 2x) / (-4x - 1)
x --> 1

=> (1(1+1)(sqrt(4) + 2) / (-4 - 1) => 2(4) / -5 => -8/5

So you mean (x^3 - x)/[√(3x+1) - 2x]? This limit satisfies the requirements for l’Hopital’s Rule.

The limit’s expression is now (3x² - 1)/[1.5/√(3x+1) – 2]

At x = 1 this expression is 2/[¾ -2] = -8/5

So the limit’s value is -1.6