Lhospitals Rule For Indeterminate Forms Homework In Spanish

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Introduction to L’Hopital’s Rule

In this article, we will describe a rule that is crucial in calculus. This rule is used to find the limits of indeterminate

forms, such as \frac{0}{0} or \frac{\infty}{\infty} . You can find different spellings of the name: L’Hopital or L’Hospital. Throughout this

article, we will use “L’Hopital.” It is important to note that this rule has nothing to do with the derivative of the quotient. L’Hopital’s rule can only be applied to rational functions (fractional). It helps us to avoid simplifying fractional functions, which occasionally can’t even be simplified.

We will present a story about the man who discovered L’Hopital’s rule, which, surprisingly, is not Guillaume François Antoine, Marquis de l’Hôpital.

Afterwards, we will demonstrate the law and give the sketch of the proof. We will also use several examples to help us understand it better.

It is essential to mention that there are situations in which L’Hopital’s rule can not or should not be applied. We will discuss frequently encountered mistakes and points of confusion, which we advise you to consider before using the method.

Who Found L’Hopital’s Rule?

L’Hopital was one of the leading French mathematicians during the 17th century. However, his name is now well-known to the world because he was lucky enough to meet his tutor, Johann Bernoulli. And guess what? Johann Bernoulli discovered L’Hopital’s Rule. The world believes that Bernoulli allowed L’Hopital to use his findings as he wished.

So, Johann Bernoulli discovered L’Hopital’s Rule.

L’Hopital’s Rule

The limit is an essential concept in calculus. It has many applications with various important results. However, it is not always straightforward to compute limits. For example, let us compute the following limits:

\lim_{x \to 0} x^2 = 0 and \lim_{x \to 0} 1-\cos{x} = 0 .

But if we are asked to calculate the limit  \lim_{x \to 0} \frac{x^2}{1- \cos{x}} , the result is not obvious. We have a similar problem when the numerator and denominator of the quotient have infinite limits, which means that the quotient does not have a determined value. These are called indeterminate forms. In such cases, L’Hopital’s Rule is the best tool to use.

There are various forms of L’Hopital’s Rule. The two considered above are widely used and easy to verify. First, let’s examine the zero over zero case.

Theorem 1:

Assume \lim_{x \to c} f\left( x \right) = 0 and \lim_{x \to c} g\left( x \right) =0 , and the functions f\left( x \right) and g\left( x \right) are

differentiable on an open interval (a,b) , which contains c .  Suppose  g^{\prime}(x) \neq 0  in  (a,b) , if x \neq c .

Then,  \lim_{x \to c} \frac{f \left( x \right) }{ g \left( x \right) } = \lim_{x \to c} \frac{f ^{\prime} \left( x \right) }{ g^{\prime} \left( x \right) } , as long as the limit is finite, \infty , or -\infty .

The result is true also for the limits x \to \infty , x \to -\infty .

In order to prove this result, we will use the Extended Mean Value Theorem. This theorem states the following:

Suppose f \left(x \right) and g \left(x \right) are differentiable on an open interval (a, b) and are continuous on the closed

interval [a, b] . Assume g^{\prime} \left( x \right) \neq 0 in [a,b] . Then, we can state that there is at least one point c in the interval  \left(a, b \right) such that \frac{f^{\prime} \left( c \right) }{ g^{\prime} \left( c \right)} = \frac{f \left( b\right) - f \left( a\right) }{ g \left( b\right) - g \left( a\right) } .

We will now prove L’Hopital’s Rule. Assume that x converges to c from the right, which we will denote as x \to c^{+} .

Let f and g be defined on an interval \left(c, b \right) , where \lim_{x \to c^{+}} f \left( x \right) = 0 and \lim_{x \to c^{+}} g \left( x \right) = 0 ,

but the ratio \frac{f^{\prime} \left( x \right) }{ g^{\prime} \left( x \right)} has a finite limit L. Then, f and its derivatives exist on some set (c, c+g] and  g^{\prime} \neq 0 on (c, c+h] . Here, we define f \left(c\right) = g \left(c\right) =0 .

According to the Extended Mean Value Theorem, there exists c_h \in (c, c+h) , such that

\frac{f^{\prime} \left( c_h \right) }{ g^{\prime} \left( c_h \right)} = \frac{f \left( c+h \right) - f \left( c\right) }{ g \left( c+h \right) - g \left( c \right) } = \frac{f \left( c + h \right) }{ g \left( c + h \right)} .

Then, when h tends to 0 from right,

\lim_{h \to 0^{+}} \frac{f^{\prime} \left( c_h \right) }{ g^{\prime} \left( c_h \right)} = \lim_{x \to c^{+}} \frac{f^{\prime} \left( x \right) }{ g^{'} \left( x\right)} .

On the other hand,

\lim_{h \to 0^{+}} \frac{f \left( c+h \right) }{ g \left( c+h \right)} = \lim_{x \to c^{+}} \frac{f \left( x \right) }{ g \left( x\right)} .

This is what we wanted to prove.

Similarly, we can verify the left convergence. Then, one may use these two sides to prove L’Hopital’s rule. We took this proof from Salas and Hille’s Calculus: One Variable.

Consider another indeterminate ratio. For instance, we want to compute the limit \lim_{x \to \infty} \frac{4x + 9}{2x^2-3} .

If we calculate this limit directly, we obtain \frac{\infty}{\infty} . Here, we would use L’Hopital’s second rule.

Theorem 2:

Suppose f and g are differentiable for all x larger than some fixed number. If \lim_{x \to \infty} f \left( x \right) = \infty and

\lim_{x \to \infty} g \left( x \right) = \infty , then: \lim_{x \to \infty} \frac{f \left( x \right)} {g \left( x \right)} = \lim_{x \to \infty} \frac{f^{\prime} \left( x \right)} {g^{\prime} \left( x \right)} , so long as the limit is finite, +\infty , or -\infty .

Now, the earlier example can be solved easily:

\lim_{x \to \infty} \frac{4x + 9}{2x^2-3} = \lim_{x \to \infty} \frac{4}{4x} = \lim_{x \to \infty} \frac{1}{x} = \frac{1}{\infty} = 0 .

It is crucial to remember that we have to differentiate the numerator and denominator of the quotient separately. Here, we will not give a sketch of the proof of the second rule, as it needs more advanced analyses. Interested readers can find it in many textbooks.

Applying L’Hopital’s Rule in Examples

In this section, we discuss several examples where L’Hopital’s Rule can be applied.

Example 1: Calculate \lim_{x \to \infty} \frac{x^2}{e^{2x}} .

First, we have to check that \lim_{x \to \infty} x^2 = \lim_{x \to \infty} e^{2x} = \infty . Then, we can apply L’Hopital’s second rule:
\lim_{x \to \infty} \frac{x^2}{e^{2x}} = \lim_{x \to \infty} \frac{\frac{d}{dx} x^2}{\frac{d}{dx} e^{2x}} = \lim_{x \to \infty} \frac{2x}{2xe^{2x}} = \lim_{x \to \infty} \frac{1}{e^{2x}} = 0 .
Example 2: Compute the following limit: \lim_{x \to 0} \frac{\sin x}{x} .

First, we check that \lim_{x \to 0} sinx = 0 and  \lim_{x \to 0} x = 0 , then we can apply L’Hopital’s first rule.

Thus, \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\cos x}{1} = 1 .

Example 3: We want to find \lim_{x \to 0^{+}} x \ln x .

Here the convergence to 0^{+} ensures the existence of the logarithm. We cannot solve this problem directly, but we also cannot apply L’Hopital’s rule yet. In order to make this possible, we have to rearrange this expression to get a ratio. Therefore, we have

\lim_{x \to 0^{+}} x \ln x = \lim_{x \to 0^{+}} \frac{\ln x}{\frac{1}{x}} .

Now, we can use L’Hopital’s rule.

\lim_{x \to 0^{+}} x \ln x = \lim_{x \to 0^{+}} \frac{\ln x}{\frac{1}{x}} = \lim_{x \to 0^{+}} \frac{\frac{1}{x}}{-\frac{1}{x^2}} =\lim_{x \to 0^{+}} -x = 0

Next, we want to discuss another example where similar modifications are required in order to apply L’Hopital’s rule.

Example 4: Compute \lim_{x \to +\infty} x \left( e^{\frac{3}{x}}- 3 \right) .

We look at the limits separately in order to verify the usage of L’Hopital’s rule. So, \lim_{x \to +\infty} x = \infty , thus  \lim_{x \to +\infty} \frac{1}{x} = 0 , and \lim_{x \to +\infty } \left( e^{\frac{3}{x}}- 3 \right) = 0 .

Afterwards, by applying our rule, we obtain \lim_{x \to +\infty} x \left( e^{\frac{3}{x}}- 3 \right) = \lim_{x \to +\infty} \frac{ e^{\frac{3}{x}}- 3 }{\frac{1}{x}} = \lim_{x \to + \infty} \frac{ e^{\frac{3}{x}} \cdot \left(-\frac{3}{x^2} \right) }{- \frac{1}{x^2}} = \lim_{x \to + \infty} 3 e^{\frac{3}{x}} = 3

In the following, we demonstrate an example where we need to perform further modifications before using L’Hopital’s rule.

Example 5: Say we want to evaluate the limit \lim_{x \to 0} x^{x} . Our expression is not a ratio and can’t be evaluated by plugging in a 0 . Here, we can use a nice trick. We use the continuous function, logarithm \ln .

We consider the logarithm of the limit. That is, \ln \left( \lim_{x \to 0} x^{x} \right) = \lim_{x \to 0} \ln (x^{x}).

Moreover, \lim_{x \to 0} \ln (x^{x}) = \lim_{x \to 0} x \ln (x) = \lim_{x \to 0} \frac{\ln x}{\frac{1}{x}} .
And now, we can apply L’Hopital’s rule, resulting in \lim_{x \to 0} \frac{\ln x}{\frac{1}{x}} = \lim_{x \to 0} \frac{\frac{1}{x}}{\frac{-1}{x^2}} = \lim_{x \to 0} -x = 0.

So, in order to find the original limit, we have to modify the last result. Finally, we get \ln{ (0) } = 1.

It is important to remember that L’Hopital’s rule can only be applied to ratios. So, first, we have to modify the original expression to obtain a quotient.

Another common mistake is to confuse L’Hopital’s rule with the derivative of the ratio. Those are entirely different rules!

L’Hopital’s Rule Sometimes Fails

It can be concluded from the previous section that L’Hopital’s rule is a nice and easy way to find limits that are difficult to compute directly. Unfortunately, there are cases when it doesn’t provide a solution, so you have to be careful.

For instance, \lim_{x \to \infty} x \left( x^2 + 1 \right)^{-\frac{1}{2}} = \lim_{x \to \infty} \frac{1}{x \left( x^2 + 1 \right)^{-\frac{1}{2}}} = \lim_{x \to \infty} \frac{\left( x^2 + 1 \right)^{\frac{1}{2}}}{x}= \lim_{x \to \infty} \frac{x \left(x^2 + 1 \right)^{-\frac{1}{2}}}{1} = \lim_{x \to \infty} x \left( x^2 + 1 \right)^{-\frac{1}{2}} = \cdots

We can see that we return to the primary expression. This means, that if we apply L’Hopital’s rule several times, we may obtain functions which never converge. Actually, this limit equals 1.

Moreover, L’Hopital’s rule is only practical when the denominator of the quotient doesn’t change signs infinitely often in the neighborhood of \infty , or when the ratio of the derivatives is unbounded.

There are situations where L’Hopital’s rule does not make computing limits easier. For example, consider the following limit:

\lim_{u \to 0} \frac{\sin x^3}{\sin ^3 x} .

To sum up, we can say that you should take care of several conditions, described above, before applying the rule.

Wrapping Up L’Hopital’s Rule

In this article, we have demonstrated L’Hopital’s rule, which is an important tool in calculus. First, we stated the rules by proving the first case. Then, we presented various examples that summarize typical problems that arise in computing limits with indeterminate forms.

It is also useful to mention that there are situations where L’Hopital’s rule fails. Thus, we brought in several examples, which investigate these issues.

So, if applied correctly and with a little care, L’Hopital’s rule is a worthwhile rule for calculus.

Let’s put everything into practice. Try this Single Variable Calculus practice question:

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