Find Limit of F(x) at x=1: 2x^2, 3, x+

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The limit of the function F(x) at x=1 is determined to be 2. The function is defined as F(x) = 2x^2 for x < 1, F(1) = 3, and F(x) = x + 1 for x > 1. To find the limit as x approaches 1, both the left-hand limit (lim(x→1-) = 2) and the right-hand limit (lim(x→1+) = 2) must be equal. Since both limits converge to the same value, the overall limit exists and is confirmed to be 2.

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F(x)= 2x^2, x<1
3, X=1
X+, x>1

Find

lim(x-->1) f(X)

f(1)=

not exactly sure how to do this. would not f(1) be just 3, since it is defined as that in the function. I am not sure about how to take that limit.

Thanks
 
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When it comes to the limit of a function at x=a, the value of the function at a is irrelevant. In fact the function may not even be defined at "a", but the limit could still exist. So the fact that f(1)=3 is irrelevant to the problem. When looking for the limit, we must find a number that the function gets very close to as x approaches "a". It must get close to the same number when approaching from the left or right, or else the limit does not exist. In the case of this function, I think you might have made an error in typing it when you said that F(x) is equal to "X+" for x>1. But whatever it is, just see what number the function gets close to as x gets close to 1 from the right and the left. If they are the same number, then this is the limit; if not, then the limit does not exist.
 
sorry that should be x+1
 
ok so

lim(x->1+)=(1+1)=2
lim(x->1-)=(1+1)=2 so the limit exists, i don't understand how that helps in finding the overall limit of F(x) at 1.
 
So the limit is 2. The limit is just the number the function approaches from both the left and the right.
 

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