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Infinite Limits; Vertical Asymptotes - Problem 3
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To find the vertical asymptote of a function, find where x is undefined. For the natural log function f(x)=ln(x), the graph is undefined at x=0. When calculating the value of the function as it gets closer and closer to 0, observe that it becomes more and more negative, so the limit as x approaches 0 is negative infinity. So, to find where any natural log function is undefined, find at what x it is equal to 0, and calculate the value of the function as it gets closer to that x to see if it approaches positive or negative infinity.

Let's do a harder problem. Consider L(x) equals natural log of 3 minus x over x plus 1. Now first, I want to find the domain of this function just a little review. Remember that for any kind of logarithm, the input has to be positive. That means 3 minus x over x plus 1 has to be greater than 0.

Now to determine whether this rational function is positive, I'm going to make a sign chart. On the sign chart, I plot points where either the numerator or the denominator equals 0. So 3, and -1. At 3, the numerator is 0, so the whole expression has a value 0. At -1, the denominator is 0, so the whole thing is undefined. All you have to do is test points in each of these three intervals to determine whether it's positive or negative.

So let's try -2. 3 minus -2, 5. -2 plus 1, -1. Positive over negative is going to be negative. So this part is not going to be in the domain. Let's go over here and try 4. 3 minus 4, -1. 4 plus 1 is 5. A negative over a positive is negative. So the domain does not include these numbers either.

Let's try something between -1, and 3. How about 0? 3 minus 0 is 3. 0 plus 1 is 1. Positive over positive. This is where the rational expression is positive. So the domain of our log function; remember, the reason we're looking at this rational function, is the inside part has to be positive. The domain is all x between -1, and 3.

So that leads us to the part B. Find the limit as x approaches 3 from the left of L(x). What happens as x approaches 3 from the left to this function? Well, let's first look at what happens to the inside part. Remember L(x) is ln of 3 minus x over x plus 1. What happens to this rational function, as x approaches 3 from the left? Well 3 minus x over x plus 1. As x approaches 3 from the left, it looks like the numerator is approaching 0. The denominator is approaching 4. So this whole thing is getting closer, and closer to 0.

Now we also know that it's positive as it's approaching 3 from the left. So it's approaching 0, but it's approaching from the positive direction. Now here we need to know a little bit more about natural log. If the input is getting closer to 0, from the positive direction, what does natural log do?

Here is a graph of natural log. As the inputs get closer to 0, as x gets closer to 0, what happens to the graph? It goes out to negative infinity. So as this goes to 0, ln of 3 minus x over x plus 1 goes to negative infinity. So that's our limit, negative infinity. Of course that indicates that there is a vertical asymptote for the graph of this function, at x equals 3. Now, there maybe other vertical asymptotes, but, the only one I have shown is x equals 3.

So just remember when you have a function that's a composition of two functions, look inside, and see what the inside function does first. Then, you can determine what the outside function does.

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