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

1 Answer
Andrea S.
Apr 17, 2018

#lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) =4#

Explanation:

Multiply and divide the function by:

#(sqrt(x^2+1)+1)/(sqrt(x^2+16)+4)#

and use the algebraic identity:

#(a-b)(a+b) = a^2-b^2#

to have:

#lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) = lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) xx (sqrt(x^2+1)+1)/(sqrt(x^2+16)+4) xx (sqrt(x^2+16)+4)/(sqrt(x^2+1)+1)#

#lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) = lim_(x->0) (x^2+1-1)/(x^2+16-16) xx (sqrt(x^2+16)+4)/(sqrt(x^2+1)+1)#

#lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) = lim_(x->0) x^2/x^2 xx (sqrt(x^2+16)+4)/(sqrt(x^2+1)+1)#

#lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) = lim_(x->0) (sqrt(x^2+16)+4)/(sqrt(x^2+1)+1)#

#lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) = (sqrt(16)+4)/(sqrt(1)+1)#

#lim_(x->0) (sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) = 4#

graph{(sqrt(x^2+1)-1)/(sqrt(x^2+16)-4) [-10, 10, -5, 5]}

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