find the limit as x approaches 1 (2-x)/(x-1)^2

the answer is infinity, but why is it that?

can someone explain?

Question

find the limit as x approaches 1 (2-x)/(x-1)^2


the answer is infinity, but why is it that? 

can someone explain?

hartnn · Answer

simply put x=1
what u get?

anonymous · Answer

1/0, undefined.

hartnn · Answer

thats correct. the limit value = 1/0 = infinity
or in other words, the limit is undefined.

anonymous · Answer

ohh okay :) thanks!

hartnn · Answer

welcome :)

anonymous · Answer

something else needs to be said, because $\frac{1}{0}$ is not infinity, even in the limit

anonymous · Answer

the reason it is $+\infty$ instead of say $-\infty$ or neither is because the denominator and numerator are both positive as $x$ gets close to zero
that is because the numerator is close to 1, and the denominator is positive, since it is a square

anonymous · Answer

I'm confused?

hartnn · Answer

did u get that denominator is always positive?

anonymous · Answer

it needs to be positive or the '0' is positive?

hartnn · Answer

since it is a square of a number and a square is always positive thats why...
now do u know why numerator is always positive ?

anonymous · Answer

because x is positive?

hartnn · Answer

because x is very near to 1 ..... like  0.999999 or 1.00000001
so 2-x must be positive.. ok ?
hence it will be +infinity and not -infinity.

anonymous · Answer

okay, but the right work to solve the prob is :

2-1/(1-1)^2=1/0=inf?

hartnn · Answer

yup, to be more accurate, just add + sign
+inf.

anonymous · Answer

it is positive because if $x$ is any number not including zero, $(x-1)^2>0$

anonymous · Answer

ohh okay

anonymous · Answer

and don't forget that when you are taking a limit, you are not plugging the number in. that is true no matter what you get. the limit $\lim_{x\to a}$  automatically assumes that $x$ is NOT $a$ but rather is close to $a$

anonymous · Answer

so the denominator is not zero, but close to zero, and it is positive because it is a square. this would not be the case for example for 
$$\lim_{x\to 1}\frac{x+1}{x-1}$$

anonymous · Answer

in that case the limit from the right would be positive infinity, but the limit from the left would be negative infinity, so the limit does not exist

anonymous · Answer

so plugging in 1 for x is the incorrect way to solve it? what is the correct way then?