How do you find the limit of #(sqrt(1+2x))-(sqrt(1-4x)] / x# as x approaches 0?

1 Answer
Jim H
Mar 28, 2016

I think there may be an error in the typing of this question. Please see the explanation section below.

Explanation:

If you want #lim_(xrarr0)(sqrt(1+2x)-sqrt(1-4x)/x)#

#lim_(xrarr0)sqrt(1+2x) = 1# and

#lim_(xrarr0)sqrt(1-4x)/x # does not exist, because as #xrarr0^+#, we have #sqrt(1-4x)/x# increases without bound.

Therefore, the difference, #lim_(xrarr0)(sqrt(1+2x)-sqrt(1-4x)/x)# does not exist because as #xrarr0^+#, the value of the expression decreases without bound.

If you want #lim_(xrarr0)(sqrt(1+2x)-sqrt(1-4x))/x# then "rationalize" the numerator.

#(sqrt(1+2x)-sqrt(1-4x))/x = ((sqrt(1+2x)-sqrt(1-4x)))/x * ((sqrt(1+2x)+sqrt(1-4x)))/((sqrt(1+2x)+sqrt(1-4x)))#

# = ((1+2x)-(1-4x))/(x(sqrt(1+2x)+sqrt(1-4x)))#

# = (6x)/(x(sqrt(1+2x)+sqrt(1-4x)))#

# = 6/(sqrt(1+2x)+sqrt(1-4x))#

Now we have

#lim_(xrarr0)(sqrt(1+2x)-sqrt(1-4x))/x = lim_(xrarr0)6/(sqrt(1+2x)+sqrt(1-4x))#

#> 6/(sqrt1+sqrt1) = 3#

Related questions
See all questions in Determining Limits Algebraically
Impact of this question
8662 views around the world
You can reuse this answer
Creative Commons License