# For what values of x, if any, does #f(x) = 1/((9x-5)sin(pi+(6pi)/x) # have vertical asymptotes?

To compute the verticle asymptotes we must consider the values for what the function is undefined at, hence where denominator

To find the second....

Using our knowledge of general solutions...

If

The graph...

Interesting...

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The function f(x) has vertical asymptotes at values of x where the denominator of the function becomes zero. In this case, the denominator is (9x-5)sin(pi+(6pi)/x). To find the values of x that make the denominator zero, we set it equal to zero and solve for x. However, sin(pi+(6pi)/x) will never be zero for any value of x. Therefore, the function f(x) does not have any vertical asymptotes.

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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.

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.

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.

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- How do you find the limit of #(3x+1/x) - (1/sinx)# as x approaches 0 using l'hospital's rule?
- How do you find the limit of #ln ( (x+1) / x )# as x approaches infinity?

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