 Another indeterminate form is something when we have a limit of a complex fraction over something else. So here we have the limit is actually going to 2 of 1 over 2x plus 6 minus 1 over 10 over x minus 2. And again, a little bit of analysis helps if we let x get close to 2. Then our numerator expression, 1 over 2 times 2 plus 6, 1 tenth minus 1 tenth, and that's going to get close to 0. Our denominator x minus 2 gets close to 2 minus 2 also gets close to 0. And that tells us that this particular limit is a 0 over 0 indeterminate form, which is actually a good thing because it means that we can do some algebraic simplification. Now what that simplification is going to be, well this is actually straight out of algebra. Anytime we have a fraction over a fraction and we can get rid of all of these internal denominators by multiplying through by the product of all the denominators present. So here we're going to multiply by the product of the two denominators that are causing us difficulties. That's 2x plus 6, this denominator up here, and this other denominator 10 right there. So I'll multiply numerator and denominator by that same quantity, and I'll do a little bit of expansion, numerator times numerator expression, denominator times denominator expression. Note that we'll keep the denominator of the big fraction in factored form because that's the most useful. And we've multiplied both of these fractions up here by this product of the denominators 2x plus 6 times 10. So here's first fraction times 2x plus 6 times 10, here's our second fraction times 2x plus 6 times 10. Now I'll do a little bit of algebra, the 2x plus 6 is here cancel, the 10's here cancel, and I end up with this, watch this step, the most common algebraic mistake here occurs when you forget that this minus here has to be distributed among both 2x and 6. That's 10 minus 2x minus 6, and the dust settles 4 minus 2x. Now we've done all this algebra so let's see if we've gotten any place. As x gets close to 2, the denominator, well because it has a factor of x minus 2, it's going to go to 0. The numerator, as x gets close to 2, 4 minus 2x gets close to 0, and both numerator and denominator both go to 0, and so we still have a 0 over 0 form. We've done a lot of work and it seems like we've gotten nowhere, except the fact that x equals 2 makes both of the numerator and denominator equal to 0, so they both have a factor of x minus 2. So I can look at the denominator and well here's my factor of x minus 2, so I don't have to worry about that. But I also know the numerator must have a factor of x minus 2, which means I can rewrite it, 4 minus 2x is x minus 2 times something, and I have to figure out what that something is, and after staring at it for a little while, I might conclude that that something is going to be negative 2. So I can replace the numerator, negative 2 times x minus 2, and guess what? We have this nice cancellation, the x minus 2's drop out, and now as x gets close to 2, I don't have denominator going to 0, so I have a chance of taking this limit. So as x gets close to 2, that's going to be negative 2, 2 times 2 plus 6 over 10, and after all the dust settles, we end up with minus 2 over 100 as by limit. Now we've done a lot of algebra and if it's possible, it's almost always worth doing a little bit of a numerical check, and so as x gets close to 2, I'm going to look at an x value close to 2, and I claim that this thing should be close to minus 2 over 100. Well, let's check it out. At x equals 1.99, I'll substitute those in, I'll evaluate, and after all the dust settles, we get negative 0.02004 approximately, and that is fairly close to our claim limit. Now, again, the actual finding of the limit, this is the important part of that problem. If you want to answer the question, find the limit, this is really the only acceptable way of finding the limit. The numerical check is a check on our answer. It's a statement that says our answer here, that we did all of this work on, is actually probably the correct answer, but the actual solution to the problem, find the limit, involves all these algebraic steps.