Please help with calculating this limit?

If you have to ask - You ought to first be plugging in numbers for "x", closer and closer but not equal to 8, to see what is happening.

Letting: F(x) = (x - 8) / (x^(2/3) - 2)

    F(7) = - 0.6  and  F(9) = + 0.4

    F(7.9) = - 0.05  and  F(8.1) = + 0.05

    F(7.99) = - 0.005  and  F(8.01) = + 0.005

    F(7.999) = - 0.0005  and  F(8.001) = + 0.0005

As "x" gets closer to 8, what number does F(x) get closer to?

For an elementry first beginner course, that will suffice. But a more serious student will want to PROVE that as "x" gets closer to 8, F(x) does indeed get closer to that number.

    First: for x>6, we have x²>36>27, so x^(2/3)>3 and x^(2/3)-2>1.

    In that case: |F(x)| < |x - 8|.

    Therefore, in order to have: |F(x)-0| < ε

    Simply choose "x" so that: |x-8| < Minimum(ε, 2)

Proving the limit discovered above.

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Eventually, you will learn about "continuous" points on functions. That is, points where the value of the functions IS the limit. That is, points on functions for which it IS true that:

    Lim F(x) = F(a)

    x->a

Most of your usual functions are continuous at almost all points. All polynomials, as well as polynomials of continuous functions, are continous at all real (and complex) numbers. All expontial functions with (real) positive bases, including Sine and Cosine, are continuous at all real (and complex) numbers. Non-integer powers (principal roots) are continous for all positive reals, but have trouble with complex numbers - particularly going around zero. All rational functions, and quotients of continuous functions, are continuous EXCEPT where the denominator is zero.

Now: 8^(2/3)-2 = 2 > 0, which means that an advanced student would immediately know that the given function was continuous and hence would merely compute:

    (8 - 8) / (8^(2/3) - 2)

To discover the desired limit.