Applications of Derivatives. L'Hopital's Rule (Part II)

Applications of Derivatives. L'Hopital's Rule (Part II)

Time for some examples.

Evaluate the limit of this fraction. Now, this is an indeterminate form. When you put in x equals infinity, the top is infinity and the bottom is also infinity. So it is an indeterminate form, infinity over infinity.

Accordingly, L'Hopital's Rule can be used to convert the limit of the original expression to the limit of this expression.

So, differentiating ex, you get ex. Differentiate x, you get one. The bottom here is the derivative of x; therefore, differentiate the bottom and get 2x-1.

If you put in x=infinity, the top is infinity and the bottom is infinity again; therefore, you have an indeterminant form.

Once again, you can apply L'Hopital's Rule. This time you differentiate the top expression, and the bottom expression. Next, check whether x is infinity for both expressions. Since the top expression equals infinity when x is infinity, but the bottom expression equals 2 when x is infinity, this is no longer an indeterminate form and the limit of this expression is infinity.

Get quality help now

WriterBelle

Verified writer

Proficient in: Math

4.7 (657)

“ Really polite, and a great writer! Task done as described and better, responded to all my questions promptly too! ”

+84 relevant experts are online

Hire writer

The next example is this expression one over sine x minus one over x, for x tends to zero. If you put in x equals zero here, you get infinity and infinity.

So, it is the indeterminate form of infinity minus infinity. The limit of this expression can be found by making it into a fraction. We do this by using the common denominator of these two fractions as our new denominator.

So the numerator turns into this.

When you plug x=0 into the denominator, the expression becomes zero over zero and becomes indeterminate. Applying L'Hopital's Rule to this expression gives you the answer.

And another example, to find the limit of this expression, something to the power of something.

Let us first define this whole expression to be the function f(x). If we take the logarithm of these functions, the log f(x), then the exponent comes down and we have log 1 + x over x. If we put in x equals zero, then the numerator becomes zero and denominator becomes zero.

So this one is zero over zero, so we can apply L'Hopital's Rule. We have the limit of this one, which is zero over zero, which is equal to the limit of this one.

If you differentiate the top, you get this; if you differentiate the bottom, you get that. If you put in x equals zero, then you do not get indeterminate form but one over one. The answer is one.

So this is the limit for the logarithm of f, so the limit of the original expression. This expression equals e, and we know that f(x)=e^log f. Therefore, the limit of log f is one. Therefore, the limit here is e to the first power, which gives you the number e.

That example illustrates the general method of finding limits. Suppose we want to find the limit of f(x), where f is some function. It is sometimes more convenient to find the limit of log f(x). If we can find the limit of log f(x) and it equals l, then we can recover the limit of f(x) in the following way: e to the l. Since log f has a limit of l, this answer becomes e to the l.