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

Question

5401 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-15T02:54:40

Change the way the ratio is written.

By the time this problem is assigned, I assume students have seen things like

#lim_(xrarr0)(sqrt(9+x)-3)/x# which is found by 'rationalizing' the numerator:

#lim_(xrarr0)((sqrt(9+x)-3))/x*((sqrt(9+x)+3))/((sqrt(9+x)+3)) = lim_(xrarr0)1/((sqrt(9+x)-3)) = 1/6#

In this problem, use the same trick on both the numerator and denominator

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

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

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

# = (sqrt(x+4)+2)/ (sqrt(x+1)+1)# #" "# (for #x != 0#)

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

# = (sqrt4+2)/(sqrt1+1) = 4/2=2#