 Some useful ideas for working with compound fractions. Any fraction can be multiplied by 1. One is any expression divided by itself, and if you multiply a fraction by its denominator, you can eliminate the denominator. It's also helpful to remember factored form is best, so unless you need to expand a product, it's best to leave it in factored form. So let's try to find this limit. So we note that our denominators are 3x-1 and 5, so if we multiply by 5 times 3x-1, the product of the two denominators, we can eliminate them both. So we'll take our expression, multiply by 5 times 3x-1 in numerator and denominator, and distribute. Now since we multiplied by 5 times 3x-1 to eliminate the denominators 5 and 3x-1, we should simplify the expressions with those denominators, but we'll leave the main denominator in factored form, so multiplying out our numerator and simplifying it. Now at our limit value x equal to 2, the numerator is 0, so we know it has a factor of x minus 2. In other words, it has to be x minus 2 times something, and that something works out to be negative 3. Now numerator and denominator have a common factor of x minus 2, so we can remove the common factor, and the algebraic expression is to find it x equals 2, so the limit is the function value, which will be, or let's find this limit. At x equals negative 1, the numerator and denominator are both equal to 0, and since we have a compound fraction, let's multiply through by the denominators. Now the best choice would be to multiply by x squared, which is the least common denominator, but let's use x times x squared, or x cubed, that's the product of the denominators. To emphasize an important point, you don't have to make the best choice as long as you make a choice, so multiplying numerator and denominator by x cubed. And again, the purpose was to get rid of the denominators x and x squared, so we'll multiply those terms out, but we'll leave the denominator with the main fraction in factored form. And to simplify our rational expression more, the numerator has a factor of x, so we could remove it, and remove the common factors, and our equality exists as long as x is not equal to 0. Now since we're concerned about the limit as x approaches negative 1, we only care about values close to x equals negative 1, so we can ignore what happens at x equals 0, and in the limit, replace the one expression with the other. So at x equals negative 1, the denominator is 0, so it must have a factor of x plus 1, and the other factor is x minus 1. We can again remove the common factor, and at x equals negative 1, the algebraic expression is defined, and so the limit is the function value, which will be...