f(x)=x^2+x-6/x-2 , what number does f(x) approach as x approaches 2?

Question

f(x)=x^2+x-6/x-2 , what number does f(x) approach as x approaches 2?

Answer 1

I don't know if you left out some parentheses.....

As written: f(x) = x^2 + x - 6 / x-2

This means that only the 6 is divided by the term x-2. If that is correct, then the last term simply approaches infinity as x approaches 2. This means the whole expression approaches infinity.

But---if you meant this:

f(x) = ( x^2 + x - 6 ) / ( x - 2 )

then you can factor the top expression to get ( x - 2 )*( x + 3 ). now, the fraction simplifies to ( x + 3 ). As x approaches 2, the value is 5

Use an online plotting utility to graph the equation---this will help you visualize. The one I like is here:

https://www.desmos.com/calculator

Answer 2

Actually the previous answer is not correct. At x=2 the function has a vertical asymptote. So the lim doesn't exists

Answer 3

limf(x)=Lim x^2+x-6/x-2 =lim((x-2)(x+3))/(x-2)=lim(x+3)=5, at x=2 the function has a hole, but it's limit exists and equals to 5