Verified answer

lim (x→∞) x/√(x²+2)

= lim (x→∞) 1/(x/√(x²+2))

= lim (x→∞) √(x² + 2)/x

= lim (x→∞) (x/√(x² + 2))/1

= lim (x→∞) x/√(x² + 2)

Which is back to the start...

When you get to:

= lim (x→∞) √(x² + 2)/x instead of using L'hopitals again do this:

= lim (x→∞) √(x² + 2)/√x²

= lim (x→∞) √(1 + 2/x²)

= √(1 + 0)

= 1

L'hopital's rule is very useful, but a lot of times it can be avoided.

Intuitively, when you see the denominator as (√x²+2), you can see that it looks like (√x²) = x. So, the limit should be equal to 1.

Let's see if our intuition is right :

x/(√x²+2)

Because you have (x^2 + 2) under the radical, and not x^2, you can't reduce (√x²+2), but you can write it like this:

(√x²+2) = [√x²(1 + 2/x²)] = x√(1 + 2/x²)

You factored out an x², so you had:

[√(x²)(1 + 2/x²)] = √(x²)√(1 + 2/x²) = x√(1 + 2/x²)

Hence, you have:

(x)/(√x²+2) = x/[(x√(1 + 2/x²)] = 1/√(1 + 2/x²)

Taking the limit as x is approaching infinity, you get:

1/√(1 + 0) = 1/√(1 ) = 1/1 = 1

I'm not sure if you have learnt this, but you can do the following instead:

Divide both top and bottom by x (since for n that approaches infinity, you can consider x not equal zero)

So you get x→∞ ( 1 ) / √1+ 2/x²

and so you get 1 / √1+ 0 = 1/1 = 0

You can do this to almost most cases when the numerator and denominator are both polynomials or same or different degrees.

1

Forget L'Hopitals rule { it's not relevant in this case}

(x)/√x²+2 x > 0

this can be simplified

(x)/√x²+2 = (x)/x+2 =( [x+2] -2 )/[x+2] = 1 + -2/[x+2] = 1 -2/u

letting u=x +2 , limit x--> oo == limit u--> oo

limit(u-->oo) 1 - 2/u = 1 { from below}

√x² == | x | == x for x > 0

I don't think you need L'Hopital. Pull the x out of the square root to get x/(x √(1 + 2/x²))

This is x/x 1/√(1 + 2/x²) = 1/√(1 + 2/x²)

This clearly converges to 1 in the limit