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

Question

1667 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-12-14T04:23:06

I would rewrite the quotient.

#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]}