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For what values of x, if any, does #f(x) = 1/(e^x-3x) # have vertical asymptotes?

Answer 1

See below.

So in general, vertical asymptotes occur when the denominator in the function is zero, such that it's value doesn't exist,
Therefore, we need #e^x-3x=0#

As it turns out, this is really quite hard to solve. I would use a graphical calculator or app to solve this.

But if we want to find solutions analytically then we could use the Taylor expansion for #e^x# which is #sum_{n=0}^inftyx^n/(n!)#. If we set this equal to #3x#, the more terms you take in the Taylor series, the closer you will get to the actual solution.

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Answer 2

Emilia Aguilar

The function f(x) = 1/(e^x-3x) has vertical asymptotes at x = 0 and x = ln(3).

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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.