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How do you find the limit of #(x^2-4x+23)# as x approaches 2?

Answer 1

William Abdalla

#lim_(x->2) (x^2-4x+23) = 19#

#f(x) = (x^2-4x+23)# is a polynomial and as such it is continuous for every #x in RR#

By definition for a continuous function:

#lim_(x->a) f(x) = f(a)#

So:

#lim_(x->2) (x^2-4x+23) = [x^2-4x+23]_(x=a) = (4-8+23) =19#

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Answer 2

Isabella Ackerman

To find the limit of a function as x approaches a specific value, we can directly substitute that value into the function.

In this case, we substitute x = 2 into the function (x^2 - 4x + 23):

(2^2) - 4(2) + 23 = 4 - 8 + 23 = 19

Therefore, the limit of (x^2 - 4x + 23) as x approaches 2 is 19.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.