Evaluating Limits using Algebra

[joshguyxc]

the problem is limit of (y^2+y-12)/(y^3-10+3) as y approaches 3. It is indeterminate, so you have to factor. I got (y-3)(y+3) for the numberator but cannot factor the denominator. The answer is 7/17 but I dont know how to get there. could you please help?

 

[srmichael]

the problem is limit of (y^2+y-12)/(y^3-10+3) as y approaches 3. It is indeterminate, so you have to factor. I got (y-3)(y+3) for the numberator but cannot factor the denominator. The answer is 7/17 but I dont know how to get there. could you please help?

I assume you meant \(\displaystyle \frac{y^2+y-12}{y^3-10y+3}\)

You factored the numerator incorrectly. You should have gotten (y + 4)(y - 3). Now, the denominator looks a little intimidating to factor since it is a cubic, however, we know that (x - 3) will most likely be a factor in order for you to cancel the (y - 3) term in the numerator. So divide the denominator by (y - 3) using either synthetic division or long division. Then see what you get and simplfy.

 

[HallsofIvy]

I assume you meant \(\displaystyle \frac{y^2+y-12}{y^3-10y+3}\)

You factored the numerator incorrectly. You should have gotten (y + 4)(y - 3). Now, the denominator looks a little intimidating to factor since it is a cubic, however, we know that (x - 3)

You mean, of course, (y- 3). And it is not a matter of "most likely". The reason we are looking at three is that it makes the denominator 0 and that means that y- 3 is a factor of the polynomial.

will most likely be a factor in order for you to cancel the (y - 3) term in the numerator. So divide the denominator by (y - 3) using either synthetic division or long division. Then see what you get and simplfy.