finding 2 limits. Unsure of answer.

I'm trying to find limits for 2 situations.

1. lim x->a

f(x)-f(a) if f(x)=x^2+4 gave me lim x ->a = 2a

--------

x-a

2. lim x->a

f(x)*(x^3-a^3) if f(x)=a^3 and g(x)=a^2 so I got lim x->a = 2a^2(a+1)

---------------

(x-a)*a^2

My main concern is that so far in every situation where we had a letter instead of a number we could get rid of the letters eventually and get a number.

I'm wondering if the answers are correct and if not what might be the correct answer and/or steps to get there.

Thank you.

Re: finding 2 limits. Unsure of answer.

Quote:

Originally Posted by

**Raitor**

1. lim x->a

f(x)-f(a) if f(x)=x^2+4 gave me lim x ->a = 2a

--------

x-a

What is f(a) if $\displaystyle f(x) = x^2 + 4$?

To start you out:

$\displaystyle \lim_{x \to a} \frac{f(x) - f(a)}{x - a} = \lim_{x \to a} \frac{x^2 - a^2}{x - a}$

The rest I leave up to you.

-Dan

Re: finding 2 limits. Unsure of answer.

[QUOTE=Raitor;735446]I'm trying to find limits for 2 situations.

1. lim x->a

f(x)-f(a) if f(x)=x^2+4 gave me lim x ->a = 2a

--------

x-a

Unfortunately, The internet does not always "respect" spaces. f(x)= (x^2+ 4)/(x- a). But since we cannot divide by 0, f(a) does not exist so f(x)- f(a) does not exist. Do you mean just "lim f(x) as x goes to a?. Well that also does not exist.

I **think** you mean "find $\displaystyle \lim_{x\to a}\frac{f(x)- f(a)}{x- a}$ where $\displaystyle f(x)= x^2+ 4$.

Okay, if that is true, $\displaystyle f(x)- f(a)= x^2+ 4- (a^2+ 4)= x^2- a^2= (x- a)(x+ a)$.

Quote:

2. lim x->a

f(x)*(x^3-a^3) if f(x)=a^3 and g(x)=a^2 so I got lim x->a = 2a^2(a+1)

---------------

(x-a)*a^2

What does "g(x)= x^2" have to do with this? There is no "g" in what you give.

If you mean $\displaystyle \lim_{x\to a}\frac{a^3(x^3- a^3}{(x- a)(a^2)}$ then you need to know that $\displaystyle x^3- a^3= (x- a)(x^2+ ax+ a^2)$. So this is the same as $\displaystyle \lim_{x\to a}\frac{a^3/a^2}\frac{x- a}{x- a}(x^2+ ax+ a^2)$

Quote:

My main concern is that so far in every situation where we had a letter instead of a number we could get rid of the letters eventually and get a number.

No, the answer is NOT a number, it will be a function of a.

Quote:

I'm wondering if the answers are correct and if not what might be the correct answer and/or steps to get there.

Thank you.

Re: finding 2 limits. Unsure of answer.

Quote:

Originally Posted by

**Raitor** 2. lim x->a

f(x)*(x^3-a^3) if f(x)=a^3 and g(x)=a^2 so I got lim x->a = 2a^2(a+1)

---------------

(x-a)*a^2

If $\displaystyle f(x) = a^3$ then f(x) is a constant. Same for g(x). I think you need to re-state this.

-Dan

Re: finding 2 limits. Unsure of answer.

Thank you for your quick answers.

I'll rewrite the whole thing also taking into account what you told me and I made a couple mistakes while writing down the second problem.

1. Evaluate $\displaystyle \lim_{x\to a}\frac{f(x)- f(a)}{x-a}$ where $\displaystyle f(x)=x^2+4$

I ended up doing this :

$\displaystyle \lim_{x\to a}\frac{f(x)- f(a)}{x-a}$ = $\displaystyle \lim_{x\to a}\frac{x^2-a^2}{x-a}$ = $\displaystyle \lim_{x\to a}\frac{(x+a)(x-a)}{x-a}$ = $\displaystyle \lim_{x\to a}(x+a)$ = $\displaystyle (a+a)$ = $\displaystyle 2a$

2. For $\displaystyle \lim_{x\to a}f(x) = a^3$ and $\displaystyle \lim_{x\to a}g(x) = a^2$ evaluate the limit of :

$\displaystyle \lim_{x\to a}\frac{f(x)*(x^3-a^3)}{(x-a)*g(x)}$

For this I did the following.

$\displaystyle \lim_{x\to a}\frac{f(x)*(x^3-a^3)}{(x-a)*g(x)}$ = $\displaystyle \lim_{x\to a}\frac{a^3*(x^3-a^3)}{(x-a)*a^2}$ = $\displaystyle \lim_{x\to a}\frac{a^3*(x-a)(x^2+ax+a^2)}{(x-a)*a^2}$ = $\displaystyle \lim_{x\to a}\frac{a^3*(x^2+ax+a^2)}{a^2}$ = $\displaystyle \lim_{x\to a}a(x^2+ax+a^2)$ = $\displaystyle a(a^2+2a+a^2)$ = $\displaystyle 2a^2(a+1)$

Wish I had known about this way to insert equations on the forums the first time around.

Would those answers be correct? What about the way I wrote the equations?

Thank you again!