Find lim (with x�¨�‡ under it) (x^2-1)/x or state that the limit does not exist.
I'm so lost with what to do with limits...I tried to simplify the expression to lim x+lim -1/x, but from there I'm pretty dumbfounded.
If anybody could help me out I'd really appreciate it.
I guess symbols do not post, but it is just lim with (x->infinity) underneath
You are looking for the limit of
x^2 -1
-------
x
as x ---> oo
well when x is very large the numerator is x^2 and the one does not amount to a hill of beans so what you really have is
x^2
---
x
which is of course x
which goes to infinity as x goes to infinity
To find the limit of the expression (x^2-1)/x as x approaches a certain value L, you need to simplify the expression as much as possible.
First, let's simplify the expression (x^2-1)/x:
(x^2-1)/x = (x+1)(x-1)/x.
Now, to find the limit, you have to consider what happens to the expression as x approaches the value L. You need to evaluate the two parts (x+1) and (x-1) separately.
To evaluate the limit of (x+1), you simply substitute the value L into the expression. So, (x+1) becomes (L+1).
To evaluate the limit of (x-1)/x, you divide both the numerator and the denominator by x:
(x/x - 1/x) = (1 - 1/x).
Now, when x approaches L, the value of 1/x approaches zero (since x approaches a finite value L but not zero), so the second part of the expression becomes 1 - 0 = 1.
Putting it all together, the limit of (x^2-1)/x as x approaches L is:
Lim (x^2-1)/x = (L+1)(1) = L + 1.
Therefore, the limit of the given expression exists and is equal to L + 1.