Evaluate the lim as x->1 of [(x^9000) - 1]/(x-1) HELP PLEASE!!!?

Answer 1

Maybe I'm missing something here, but I think you're supposed to use L'Hopital's rule.

Taking the derivative of the top and bottom yields 9000x^8999, substitute in x and you'll get 9000.

Answer 2

If you DO the division, which you can if you're not dividing by 0, you get x^8999 + x^8998 + ... + x² + x + 1, and if x = 1, that's 9000. How do you handle lim x→2 [(x-2)(x+3)]/(x-2) ? The graph is a line with a hole at (2, 5), where 5 is the value of x+3 when x=2. You have a removable discontinuity, so the limit at 2 is 5. Same here. Lim is 9000. On a graphing calculator, use the table feature, let table start be 1, delta = 0.00001, and see what's up.

Answer 3

Evaluate the limit as x→1 of [(x^9000) - 1]/(x-1)

This is of indeterminant form 0/0 so we can apply L'Hospital's rule which says the the limit of the quotient is equal to the quotient of the derivative of the numerator divided by the derivative of the denominator.

So we have

limit as x→1 of [(x^9000) - 1]/(x-1)
= limit as x→1 of [9000(x^8999)]/1 = 9000*1 = 9000

Answer 4

Divide the numerator by the denominator.

You get 9000 terms added together, each of which has limit 1.

So the limit is 9000.

Answer 5

[(x^9000) - 1]/(x-1)

=x^8999+x^8998+x^8997+ ... + x^2 + x + 1

= 9000 (lim x -> 1)