a) What is wrong with the equation (x^2+x-6)/x-2=x+3 b) In view of part a explain why the equation is correct lim x->2(x^2+x-6)/x-2=lim x->2(x+3)

Question

I get x + 3 = x + 3. What am I not seeing here?

Answer 1

There is a reason they are called equivalent fractions an not equal.

let f(x) = (x^2 + x -6)/(x -2) and let g(x) = x + 3. 

 

These are not equal functions because f(2) is undefined while g(2) = 5. For every other value, these 2 functions are equal, but they are not always equal.

 

(x^2+x-6)/(x-2) reduces down to x+3 but they are not ALWAYS equal.

 

The graphs of these are therefore the same except when x =2. f(x) is the line represented by g(x) except it has a hole in it. This is called a removable discontinuity. As x is very close to 2, lets say 2.000001, f(2.000001) = g(2.000001). You can get as close to 2 as you want and these functions will always be equal. At the actual value x =2, these 2 functions are not equal.

 

Hope this makes sense.

Answer 2

there is nothing wrong with the equation as long as you state the domain as all real numbers except x = 2 because the expression has no value at x=2

 

if we say f(x) = (x² + x - 6)/(x + 3), f(2) is undefined but the limit f(x) as x -> 2 equals 5

 

if a function is continuous at c, then f(c) = lim f(x) as x->c

 

when f(c) is undefined, there is a discontinuity at c

 

I think the mistake many students make is thinking that if the limit of f(x) as x ->c has a value L, then f(c) = L.

 

This is true if the function is continuous but in general these two things should be thought of as two different expressions.