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How do you find #lim (x^2+4)/(x^2-4)# as #x->2#?

Answer 1

There is no such limit.

The function has vertical asymptotes at x =-2 and x = +2. Informally, #x to oo# as #x to 2#.

Furthermore, as #x to +-oo#, the function #to# 1, because the coefficients of the highest power of #x# in the numerator polynomial is the same as that in the denominator polynomial.

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Answer 2

William Abdalla

To find the limit of (x^2+4)/(x^2-4) as x approaches 2, we substitute the value of 2 into the expression. However, this results in an undefined expression since the denominator becomes zero. Therefore, we need to simplify the expression before substituting. By factoring the denominator as (x+2)(x-2), we can cancel out the common factors in the numerator and denominator. After canceling, we are left with (x+2)/(x-2). Now, we can substitute 2 into this simplified expression, which gives us (2+2)/(2-2) = 4/0. Since division by zero is undefined, the limit does not exist.

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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.

How do you find #lim (x^2+4)/(x^2-4)# as #x-&gt;2#?

How do you find #lim (x^2+4)/(x^2-4)# as #x->2#?