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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 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,404 Thanks: 2477 Math Focus: Mainly analysis and algebra Start by remembering the definition of a continuous function. How could you apply this to each of the questions?
 January 25th, 2018, 06:53 AM #3 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 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 $8x^3- 32x^2= 8x^2(x- 4)$.
 January 25th, 2018, 08:20 AM #4 Newbie   Joined: Jan 2018 From: MI Posts: 11 Thanks: 1 Thanks, I figured it out and got the answers! Thanks from Joppy
February 12th, 2018, 03:53 PM   #5
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Math Focus: Yet to find out.
Quote:
 Originally Posted by Mathlover1234 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: https://youtu.be/ISNb73yg0Qg
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