How do you find the asymptotes for #ln(x-2)#?

Answer 1

Vertical asymptote at #x=2.#

A logarithmic function has a vertical asymptote at #x=c# where #c# is the value of #x# causes the argument inside the parentheses to become #0.# This is because #log_a(x), ln(x)# do not exist for #x<0.#

For #ln(x-2):#

#x-2=0#

#x=2#

Is the vertical asymptote, as for values less than #x=2, ln(x-2)# doesn't exist.

As for horizontal asymptotes, logarithmic functions don't generally have any (at least in the forms #log_a(x), ln(x).#)

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Answer 2

Lily Anderson

To find the asymptotes for ( \ln(x-2) ), you need to consider the domain of the natural logarithm function. Since the natural logarithm is defined only for positive values, the expression ( x - 2 ) inside the logarithm must be greater than zero. Thus, the domain of ( \ln(x-2) ) is ( x > 2 ).

Therefore, there are no vertical asymptotes for ( \ln(x-2) ). However, there is a vertical asymptote at ( x = 2 ) since the function is undefined for ( x = 2 ).

In summary, the only asymptote for ( \ln(x-2) ) is the vertical asymptote at ( x = 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.