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

##### 1 Answer

Dec 14, 2016

I would rewrite the quotient.

#### Explanation:

# = 4/sqrt2 = 2sqrt2#

**Note that**

For

Since

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

So the limit as

Here is the graph:

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