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January 25th, 2018, 06:22 AM   #1
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Calc Help

I can't seem to figure out the answer for both these questions. Thank you
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January 25th, 2018, 06:52 AM   #2
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Start by remembering the definition of a continuous function.

How could you apply this to each of the questions?
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January 25th, 2018, 06:53 AM   #3
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Do you know the basic definitions? A function, f, is "continuous at x= a" if and only if $\displaystyle \lim_{x\to a} f(x)= f(a)$.

What is the limit, as x goes to 4, of 4x? What is the limit, as x goes to 4, of $2x^2$? In order that the limit of f exist at x= a, those must be the same.

The other one is basically the same, perhaps a little harder. What is the limit, as x goes to 4, of $\frac{8x^3- 32x^2}{x- 4}$? It helps to notice that $8x^3- 32x^2= 8x^2(x- 4)$.
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January 25th, 2018, 08:20 AM   #4
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Thanks, I figured it out and got the answers!
Thanks from Joppy
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February 12th, 2018, 03:53 PM   #5
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Originally Posted by Mathlover1234 View Post
So I just stumbled upon this proof as to why 0!=1. Pretty cool stuff.

If n! is defined as the product of all positive integers from 1 to n, then:
1! = 1*1 = 1
2! = 1*2 = 2
3! = 1*2*3 = 6
4! = 1*2*3*4 = 24
n! = 1*2*3*...*(n-2)*(n-1)*n
and so on.
Logically, n! can also be expressed n*(n-1)! .

Therefore, at n=1, using n! = n*(n-1)!
1! = 1*0!
which simplifies to 1 = 0!

Also watch this cool video:
Why are you posting this here?
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