Question

How do you find `lim (4x)/sqrt(2x^2+1)` as `x->oo` using l'Hospital's Rule or otherwise?

Answer 1

I would rewrite the quotient.

Explanation:

`lim_(xrarroo)(4x)/sqrt(2x^2+1) = lim_(xrarroo)(4x)/(xsqrt(2+1/x^2)`

`= 4/sqrt2 = 2sqrt2`

Note that

For `x != 0`, we have `sqrt(2x^2+1) = sqrt(x^2)sqrt(2+1/x^2)`

Since `sqrt(x^2) = absx`, and we are interested in the limit as `x` increases without bound, we have `abs(x^2) = x`

For #`x`# decreasing without bound, we get

`sqrt(x^2) = -x`.

So the limit as `xrarr-oo` would be `-2sqrt2`.

Here is the graph:

graph{(4x)/(sqrt(2x^2+1) [-16.01, 16.02, -8.01, 8]}