Solve Calculus Problems: Limits and Gradients Explained | Algebra II Tips

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In summary, for the first problem, the limit is 1/12 and for the second problem, the gradient of a function with an exponent and a variable can be found using the formula y'= ln(a) ax.
  • #1
Sombra
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Ok, I have 2 problems. I think my first one deals with an algebra II thing. But, it's: Evaluate the limit (if it exists):

lim (x approaches -2) (x+2)/(x^3 +8)

I think my problem came from factoring the denominator, which I got as
(x+2)(x^2 - 4x +2), so my answer came out to be 1/14, which is not one of the answer choices. I can't remember factoring cubes. I know it's SOPSMS, but I can't even remember how to do that.

My other problem. Can anyone give me an example of finding the gradient of a function at, say x=3, for something that has an exponent with a variable? Maybe 4^x. Thanks!
 
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  • #2
You cube is almost correct but you needed to multiply 2 and x and have 2^2 at the end so you should get (x+2)(x^2 - 2x +4). The answer is 1/12
 
  • #3
Sombra said:
My other problem. Can anyone give me an example of finding the gradient of a function at, say x=3, for something that has an exponent with a variable? Maybe 4^x. Thanks!

I assume you want, for example:
[tex]\frac{d}{dx} 4^x=\ln 4 \times 4^x[/tex]

(It's usually called a derivative in english. Gradient refers to other things.)
 
  • #4
In a bit more detail: if y= ax, then ln(y)= x ln(a) so (ln(y))'= (1/y)y'= ln(a)
Then y'= ln(a) y= ln(a) ax.
 

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