How do you calculate this limit without using l’Hospital’s rule: lim √(x^2 +1) -1 / √(x^2+16) - 4 as x → 0 ?

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Answer 1 · 2018-04-15T11:19:54

Please see below.

#"Your "color(red)"Question:"#

#color(red)(√(x^2 +1) -1 / √(x^2+16) - 4#

#"Answer for the "color(blue)" Question:"#

#color(blue)(lim_(xto0)(sqrt(x^2+1)-1)/(sqrt(x^2+16)-4)#

#lim_(xto0) [(sqrt(x^2+1)-1)/(sqrt(x^2+16)-4)xx(sqrt(x^2+1)+1)/(sqrt(x^2+1)+ 1)xx(sqrt(x^2+16)+4)/(sqrt(x^2+16)+4)]#

#=lim_(xto0) [((x^2+1)-1)/(sqrt(x^2+1)+1)xx(sqrt(x^2+16)+4)/((x^2+16)-16)]#

#=lim_(xto0) [((cancelx^2)/(sqrt(x^2+1)+1)xx(sqrt(x^2+16)+4)/(cancelx^2)]#

#=1/(sqrt(0+1)+1)xx(sqrt(0+16)+4)/1#

#=1/2xx8#

#=4#