How do you find the limit #(x^3+4x+8)/(2x^3-2)# as #x->1^+#?

1 Answer
Timber Lin
Dec 27, 2017

#+oo#

Explanation:

use direct substitution:

#lim_(xrarr1^(+))(x^3+4x+8)/(2x^3-2)#

#=((1^(+))^3+4(1^(+))+8)/(2(1^(+))^3-2)#
(#1^(+)# means a number slightly larger than 1, such as 1.0000001)

#=(1^(+)+4^(+)+8)/(2(1^(+))-2)#

#=(13^(+))/(2^(+)-2)#

#=(13^(+))/(0^(+))#

#=+oo#

graph{(x^3+4x+8)/(2x^3-2) [0.9, 1.1, -1000, 1000]}
from the graph, you can see that the function approaches #+oo# as x approaches 1 from the right side

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