How do you find #lim x^2/(sqrt(2x+1)-1)# as #x->0# using l'Hospital's Rule?

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Answer 1 · 2017-08-29T21:30:25

We start by checking to make sure the limit is actually of the form #0/0# or #oo/oo#.

#L = 0^2/(sqrt(2(0) + 1) - 1) = 0/(1 - 1) = 0/0#

So we may indeed use l'hospital's.

The derivative of #sqrt(2x + 1)# is #2/(2sqrt(2x + 1)) = 1/sqrt(2x+ 1)#.

#L = lim_(x->0) (2x)/(1/sqrt(2x + 1)) #

#L = lim_(x->0) 2xsqrt(2x + 1)#

If we evaluate now, we get:

#L = 2(0)sqrt(2(0) + 1) = 0#

Looking at the graph, we realize that #y# does approach #0# as #x# approaches #0#.

graph{x^2/(sqrt(2x+ 1) - 1) [-4.93, 4.934, -2.465, 2.465]}

Hopefully this helps!