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How do you find the limit #(x+5)(1/(2x)+1/(x+2))# as #x->oo#?

Answer 1

Paisley Johnson

First write it as a single ratio.

#(x+5)(1/(2x)+1/(x+2)) = (3x^2+17x+10)/(2x^2+4x)#

For #x != 0# we can factor out #x^2# in both the numerator and the denominator to get

# = (x^2(3+17/x+10/x^2))/(x^2(2+4/x))#

# = (3+17/x+10/x^2)/(2+4/x)# (for #x != 0#)

As #x# increases without bound all of #17/x#, #10/x^2# and #4/x# approach zero.

Therefore,

#lim_(xrarroo) (x+5)(1/(2x)+1/(x+2)) = lim_(xrarroo)(3+17/x+10/x^2)/(2+4/x)#

# = (3+0+0)/(2+0) = 3/2#

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Answer 2

Emily Ammerman

To find the limit of the expression (x+5)(1/(2x)+1/(x+2)) as x approaches infinity, we can simplify the expression first. By multiplying the terms, we get (x+5)/(2x) + (x+5)/(x+2).

Next, we can divide each term by the highest power of x in the denominator, which is x in this case. This gives us 1/2 + 1 = 3/2.

Therefore, the limit of the expression as x approaches infinity is 3/2.

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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+5)(1/(2x)+1/(x+2))# as #x-&gt;oo#?

How do you find the limit #(x+5)(1/(2x)+1/(x+2))# as #x->oo#?