How to find the limit at inifity of...

What steps would be best/ easiest to determine the limit at infinity of this problem:

lim x/ √(x²-2)
x→∞

Best answer gets a brownie that I'm baking as I type!

A brownie! ;)

First, multiply the top and bottom by 1/x (i.e. 1/x / 1/x * function). Since the bottom is a square root, you can change the bottom multiple to √(1/x²). You should get:

lim 1/√(1 - 2/x²)
x→∞

Now, since x→∞, the 2/x² = 0, so it becomes

lim 1/√(1 - 0)
x→∞

or

lim 1/√1
x→∞

now, you can solve directly:

1/√1 = 1/1 = 1

Basically, the reason for the first step of dividing both top and bottom by 1/x is that you want to end up with all the x's on the bottom of ratios. This is because any ratio that has infinity on the bottom (assuming infinity is also not on top), equals zero and you can remove it from the equation.

One thing we've learned to do in our calculus class is to divide everything by the largest exponent. In this case, that would be x^1, so let's do that:

lim x/√(x²-2)*(x/√x²) (√x²=x for all numbers greater than 0)
x→∞

lim 1/√(1-2/x²)
x→∞

1/√(1-0)
1/√1
1/1
1

So you're function should approach 1 as x approaches infinity. I hope that brownie doesn't have nuts, as I am allergic.

You must use L'Hopital's rule for this, since it is inf/inf, which is indeterminate.

Derive the numerator, then derive the denominator; you don't need the quotient rule for this.

You'll have to do this a couple times, and then you'll get that the limit = 1.

There's a lot of work with this problem, so have fun with it. ;)

You can sub in a very large number for x into the equation, and see what number the answer will approach

sub in x = 1000

1000 / √(1000²-2)
= 1.000001

so it approaches 1 from above

the answer is Positive infinity
because the top exponent (1) is bigger than the bottom (1/2) => 1/2 because of the square root