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How do you find the limit of #(8x -lnx)# as x approaches infinity?

Answer 1

Scarlett Allison

Use #8x-lnx = x(8-lnx/x)#.

#8x-lnx = x(8-lnx/x)#.

#lim_(xrarroo)lnx/x=0# #" "# (Use l"Hospital if you don't know this limit.)

#lim_(xrarroo)(8x-lnx) = lim_(xrarroo)(x(8-lnx/x)) = oo(8-0) = oo#.

I'm not that comfortable using this notation, but it's so convenient!

Here it is expressed without the notation.

As #x# increases without bound, #8-lnx/x rarr 8#,

so #x(8-lnx/x)# is the product of an expression whose value increases without bound and an expression whose value approaches a positive value. The product, therefore increases without bound.

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Answer 2

Charles Anctil

To find the limit of (8x - ln(x)) as x approaches infinity, we can use the concept of limits.

First, we observe that as x approaches infinity, the term 8x grows without bound.

Next, we consider the term ln(x). As x approaches infinity, ln(x) also grows without bound, but at a slower rate compared to 8x.

Therefore, the dominant term in the expression (8x - ln(x)) is 8x.

Hence, as x approaches infinity, the limit of (8x - ln(x)) is infinity.

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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.