Need help with Limits?

Question

the limit as x approaches 8

function: sin (x-8)/(X^2-2x-48)

you can view it here:
http://www.webassign.net/www28/symImages/5/1/1a1872403cd882533309a7f2a07592.gif

Thanks. If you could please explain or show work, I'd appreciate it.

the sin is only on the numerator (please click the link to see)

Answer 1

L'Hopital's Rule is the simplest way to solve this problem. But if you haven't gotten that far yet, you can solve it by a method of partial fractions.

1/(x^2-2x-48) = 1/[(x-8)*(x+6)] = A/(x-8) + B/(x+6)
1 = A*(x+6) + B*(x-8)
Plug in x = 8 shows that A = 1/14
Plug in x = -6 shows that B = -1/14

1/(x^2-2x-48) = 1/[14*(x-8)] - 1/[14*(x+6)]
sin(x-8)/(x^2-2x-48) = sin(x-8)/[14*(x-8)] - sin(x-8)/[14*(x+6)]

The limit as x approaches 8 of the left side is the sum of the limits of the left side, if they exist.

limit as x approaches 8 of the first term is 1/14 since sin(x)/x approaches 1 as x approaches 0. limit as x approaches 8 of the second term is 0.

Therefore, the limit as x approaches 8 is 1/14 + 0 = 1/14.

Answer 2

Wouldn't L'Hopital's rule work here?

Take the derivative of the numerator and denomenator and take the limit again.

cos(x-8)/(2x-2). cos(0) = 1 2*8-2 = 14.
The limit is 1/14
?

Answer 3

the easiest way to do that is make a t chart and plug values that get closer to 8 like 7.9 for x then you get a y value then plug 7.99 and then 7.99 once you know what it is approaching there is your limit (you obviously need a calculator)