How do you evaluate the limit #(x^2+2x-8)/(sqrt(x^2+5)-(x+1))# as x approaches 2?

Question

Emma Aldridge · Accepted Answer

-18

#lim_{x to 2} (x^2+2x-8)/(sqrt(x^2+5)-(x+1))# #= lim_{x to 2} ((x+4)(x-2))/(sqrt(x^2+5)-(x+1))# let #x = 2 + h, 0< abs(h) "<<" 0# so the limit becomes #= lim_{h to 0} ((6+h)(h))/(sqrt(9 + 4h + h^2)-(3+h))# by Binomial expansion #sqrt(9 + 4h + h^2)# #= 3 (1 + 1/9 (4h+h^2) )^(1/2)# #= 3 (1 + 1/2* 1/9 (4h+h^2) + O(h^2) )#, remembering that # 0< abs(h) "<<" 0# #= 3 + 2/3 h + O(h^2) # #implies lim_{h to 0} ((6+h)(h))/(3 + 2/3 h + O(h^2) -(3+h))# #= lim_{h to 0} ((6+h)(h))/(-1/3 h + O(h^2) )# #= lim_{h to 0} - (3(6+h))/(1 + O(h))# #= -18#

Charles Arocha · Answer

undefined To evaluate the limit of (x^2+2x-8)/(sqrt(x^2+5)-(x+1)) as x approaches 2, we can substitute the value of x into the expression. However, this would result in an undefined expression since the denominator would be zero. To overcome this, we can simplify the expression by multiplying both the numerator and denominator by the conjugate of the denominator, which is sqrt(x^2+5)+(x+1). Simplifying further, we get (x^2+2x-8)/(x-2). Now, substituting x=2 into this simplified expression, we get (2^2+2(2)-8)/(2-2) = 0/0. This is an indeterminate form. To evaluate this limit further, we can use algebraic manipulation or techniques such as L'Hôpital's rule or factoring to simplify the expression and find the limit.