Thread: Finding the answer to a Limit

Find lim->0 (2+5/x²)
This is what I did first.
I first plugged in 0 to find if it's true.
2+5/0
Of course it's undefined since you can't divide by 0
So I was confused... How do you do this then?

Because if you factor x², it'll be (x)(x) which will be 0 = undefined
Can someone explain please?

you don't actually divide by 0. The concept of taking a limit is going towards 0, but you have to think you will never actually reach that number. How you need to think about it is that you're dividing 2+5 by a number extremely small. So the answer is infinity. Any number divided by something infinitely small will yield an extremely huge number. Hence as you approach 0, (2+5)/x^2 tends to infinity.

Oh oops, it was actually 2+(5/x^2)
But I see, thanks

The more interesting problems occur when you get X occuring more than once,

eg Lim x->inf [(x^2+1)/x]-x = 0

Can be solved as below,

Lim x->inf = x^2/x + 1/X -X

X^2/X =X
1/X=0

Lim x->inf= X + 0 -X = 0