How do you solve #\lim _ { x \rightarrow \pm \infty } ( \frac { x ^ { 2} - 2x + 1} { x ^ { 2} + 4x + 5} ) ^ { 2}#?

1 Answer
Somebody N.
Nov 12, 2017

See below.

Explanation:

This is really two different limits, one to #oo# and one to #-oo#.

They can't be written as:

#lim_(x->+-oo)#: This make little sense. This is two different limit points.

You can have:

#lim_(x->oo)#: and #lim_(x->-oo)#:

Maybe you mean left and right limit, but this is not possible with limits to infinity. You can't be on the left or right side of infinity, because infinity is a concept not a value.

For:

#lim_(x->oo)((x^2-2x+1)/(x^2+4x+5))^2#

For limits to infinity we only need be concerned with the two highest powered terms, since these are increasing faster than lower powered term and the constants make no real contribution to the value.

So we have:

#(x^2/x^2)^1= (1)^2=1#

#:.#

#lim_(x->oo)((x^2-2x+1)/(x^2+4x+5))^2=1#

For:

#lim_(x->-oo)((x^2-2x+1)/(x^2+4x+5))^2#

This will be #(1)^2=1#

#:.#

#lim_(x->-oo)((x^2-2x+1)/(x^2+4x+5))^2=1#

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