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

Answer 1

Emilia Aguilar

#+oo#

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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Answer 2

Ezekiel Alexander

To find the limit of (x^3+4x+8)/(2x^3-2) as x approaches 1 from the positive side, we substitute 1 into the expression. This gives us (1^3+4(1)+8)/(2(1^3)-2), which simplifies to (1+4+8)/(2-2). Further simplifying, we have 13/0. Since the denominator is zero, the limit does not exist.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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

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