How do you find the limit #x^2/sqrt(2x+1)-1# as #x->0#?

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Answer 1 · 2017-01-02T16:33:54

As asked, at #x=0#, the denominator is not #0#. So use substitution.

For #lim_(xrarr0)x^2/(sqrt(2x+1)-1)# see below.

As asked

For #x# close to #0#, the numerator, #x^2# is also close to zero.

The denominator is close to #sqrt(2(0)+1) = sqrt1 = 1#

So the quotient is close to #0/1# which is #0#.

If a different quotient was intended

If the question is missing parentheses and should be x^2/(sqrt(2x+1)-1), then use

#x^2/(sqrt(2x+1)-1) = x^2/((sqrt(2x+1)-1)) * ((sqrt(2x+1)+1))/((sqrt(2x+1)+1))#

# = (x^2(sqrt(2x+1)+1))/((2x+1)-1)#

# = (x^2(sqrt(2x+1)+1))/(2x)#

# = (x(sqrt(2x+1)+1))/2#.

So, the limit is #(0(sqrt(2(0)+1)+1))/2 = 0#