Question

Can anyone tell me why by direct substitution of x, the equation (circled ones) equals to the indeterminate form, 0/0? When you can do x²-k=6 something like that​

Answer 1

See explanation and hopefully it answers your question.

Basically because the expression has a hole at x=3.

Explanation:

Let h(x)=( x^2-k ) / ( hx-15 )

This function, h, has a hole in the curve at hx-15=0 if it also makes the numerator 0 for the same x value.

Solving for x in that equation:

Adding 15 on both sides:

hx=15

Dividing both sides by h:

x=15/h

For it be a hole, you also must have the numerator is zero at x=15/h.

x^2-k=0 at x=15/h gives:

(15/h)^2-k=0

225/h^2-k=0

k=225/h^2

So if we wanted to evaluate the following limit:

Lim x->15/h ( x^2-k ) / ( hx-15 )

Or

Lim x->15/h ( x^2-(225/h^2) ) / ( hx-15 ) you couldn't use direct substitution because of the hole at x=15/h.

We were ask to evaluate

Lim x->3 ( x^2-k ) / ( hx-15 )

Comparing the two limits h=5 and k=225/h^2=225/25=9.