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

Question

Christopher Allers · Accepted Answer

For #x->2# we have that #(x-2)^2# can be made as small as we want and is always positive, so that:

#1/(x-2)^2#

can be made as large as we want and is always positive.

Then:

#lim_(x->2) 1/(x-2)^2 =+oo#

Formally, for any #M >0# choose #delta_epsilon < sqrt(1/M)#. Then for #x in (2-delta_epsilon,2+delta_epsilon)# we have that:

# abs(x-2) < delta_epsilon < sqrt(1/M)#

and so:

#(x-2)^2 < 1/M#

and:

#1/(x-2)^2 > M#

which proves the limit.

Ezekiel Alexander · Answer

undefined To determine the limit of 1/(x-2)^2 as x approaches 2, we can use algebraic manipulation. By substituting x=2 into the expression, we get 1/(2-2)^2, which simplifies to 1/0. Since division by zero is undefined, we cannot directly evaluate the limit using substitution. However, we can use a different approach called factoring. By factoring the denominator, we can rewrite the expression as 1/[(x-2)(x-2)]. Canceling out the common factors, we are left with 1/(x-2). Now, substituting x=2 into this expression, we get 1/(2-2), which simplifies to 1/0. Again, division by zero is undefined. Therefore, the limit of 1/(x-2)^2 as x approaches 2 does not exist.