How do you determine the limit of #[1/(x-2) + 1/(x+2)]# as x approaches 2+?

Question

Aurora Alvarado · Accepted Answer

The Limit in question does not exist.

To find #lim_(xrarr2+){1/(x-2)+1/(x+2)}#, we note that, As #xrarr2+, x>2, (x-2)>0, i.e., (x-2)rarr0+# #rArr1/(x-2)rarr oo# Thus, despite #lim_(xrarr2+)1/(x+2)=1/4#, we find that, #lim_(xrarr2+){1/(x-2)+1/(x+2)}# does not exist.

Ezra Adams · Answer

undefined To determine the limit of [1/(x-2) + 1/(x+2)] as x approaches 2+, we can simplify the expression by finding a common denominator and combining the fractions. The common denominator is (x-2)(x+2). After combining the fractions, we get [(x+2) + (x-2)] / [(x-2)(x+2)]. Simplifying further, we have 2x / [(x-2)(x+2)]. As x approaches 2+, we substitute the value into the expression, resulting in 2(2) / [(2-2)(2+2)]. Simplifying this, we get 4 / 4, which equals 1. Therefore, the limit of [1/(x-2) + 1/(x+2)] as x approaches 2+ is 1.

Samantha Adkison · Answer

undefined To determine the limit of the function [1/(x-2) + 1/(x+2)] as x approaches 2 from the right side (2+), we can simply substitute x = 2 into the function and evaluate it. Plugging in x = 2 into the function yields:

1/(2-2) + 1/(2+2) = 1/0 + 1/4 = ∞

Therefore, the limit of the function as x approaches 2 from the right side is infinity.