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

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Answer 1 · 2017-02-16T21:55:13

This is not an indeterminate form. See below.

As #xrarr0#, the numerator goes to #2#. Therefore, the form is not indeterminate and it would be an error to use l'Hospital's Rule.

As #xrarr0#, the denominator goes to #0#, so the limit from the right is either #oo# or #-oo#.

To determine which, consider an #x#, just a bit greater than #0#.

Then #x+1# is a bit more than #1#, and #sqrt(x+1)# is a bit greater than #1#. So the denominator is a positive number near #0#.

We can write the form of the limit as #2/0^+# which tells us that, as #xrarr0^+#, the quotient increases without bound.

We write

#lim_(xrarr0^+) (sqrt(x+1)+1)/(sqrt(x+1)-1) = oo#

Answer 2 · 2017-02-16T22:19:11

There is no limit.

Generally:

#lim_(x to 0) (sqrt(x+1)+1)/(sqrt(x+1)-1)#

#= lim_(xto 0) ((1 + 1/2x + ...) +1)/((1 + 1/2 x + ...)-1) #

#= lim_(xto 0) ((2 + 1/2x + ...) )/((1/2 x + ...)) #

For small positive #x#, this is always going to be positive. For small negative #x#, this is always going to be negative.

So there is no limit :(

We have a 2-sided limit.