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

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The discussion focuses on evaluating a limit and finding the gradient of a function. The limit problem involves evaluating lim (x approaches -2) (x+2)/(x^3 +8), where the correct factoring of the denominator is crucial for finding the right answer, which is 1/12. Additionally, the gradient of the function 4^x is discussed, with the derivative given as d/dx 4^x = ln(4) * 4^x. The conversation clarifies that "gradient" typically refers to different concepts in calculus, emphasizing the importance of terminology. Understanding these concepts is essential for solving related calculus problems effectively.
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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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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
 
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:
\frac{d}{dx} 4^x=\ln 4 \times 4^x

(It's usually called a derivative in english. Gradient refers to other things.)
 
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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