Home > Calculus > Infinite Limits and Vertical Asymptotes

For what values of x, if any, does #f(x) = 1/((x+2)(x-1)) # have vertical asymptotes?

Answer 1

#x=-2# and #x=1#

Vertical asymptotes will occur whenever the denominator of the function is equal to #0#.

So, the vertical asymptotes are at the values of #x# that satisfy

#(x+2)(x-1)=0#

Just like when solving a quadratic equation, we set both of these terms equal to #0# individually. When terms are being multiplied by one another to equal #0#, at least one of the terms must also equal #0#.

So, in setting both term equal to #0#, we find the #2# values where there are vertical asymptotes:

#x+2=0" "=>" "color(red)(x=-2#

#x-1=0" "=>" "color(red)(x=1#

We can check a graph of the function:

graph{1/((x+2)(x-1)) [-5, 5, -5, 5]}

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Answer 2

Evelyn Agard

The function f(x) = 1/((x+2)(x-1)) has vertical asymptotes at x = -2 and x = 1.

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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.