While L'Hopital's Rule is a solid technique for evaluating limits that turn into \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) with direct substitution, with some creative thinking it can be used to evaluate other limits as well. In this article, we look for more creative ways to apply the rule, with the theme being attempts to force usage of L'Hopital's Rule by creatively manipulating the original limit. Note that this article assumes a basic familiarity with L'Hopital's Rule, so read other articles provided if you are not yet at this level, before continuing forward.
Arbitrary Constants
Sometimes we can force L'Hopital's Rule to apply to a limit by combining it with a constant or another limit. In general, we set the value of the original limit as \(c\), a constant, and then use that equation to manipulate the limit into a form with which it is easy to find \(c\). This technique is one best demonstrated by example.
Example 1: Find the value of the limit \(\lim_{x \rightarrow \infty}\frac{4}{x^2}\).
Solution 1: Let this limit be equal to an arbitrary constant, so
$$\lim_{x \rightarrow \infty}\frac{4}{x^2} = c$$
Direct substitution makes this limit of the form \(\frac{4}{\infty}\), which does not allow the use of L'Hopital's Rule. However, we can add this limit to another one with a known value to make the resulting limit a \(\frac{\infty}{\infty}\) indeterminate form. Add \(1\) to both sides of the equation:
$$\lim_{x \rightarrow \infty}\frac{4}{x^2} + 1 = c + 1$$
We can include the \(+1\) on the left side of the equation within the limit--this is a fundamental rule of limits; we can extend the piece of an expression that a limit applies to. Thus we have
$$\lim_{x \rightarrow \infty}\left(\frac{4}{x^2} + 1\right) = c + 1$$
Now simplify the limit so that it is simply one fraction:
$$\lim_{x \rightarrow \infty}\left(\frac{4}{x^2} + \frac{x^2}{x^2}\right) = c + 1 \Rightarrow$$
$$\lim_{x \rightarrow \infty}\frac{4 + x^2}{x^2} = c + 1$$
Now the limit is an \(\frac{\infty}{\infty}\) indeterminate form! Use L'Hopital's Rule on it:
$$\lim_{x \rightarrow \infty}\frac{2x}{2x} = c + 1 \Rightarrow$$
$$1 = c + 1$$
The limit is gone! We defined the value of the original limit to be \(c\), so solving this equation for \(c\) gives us our final answer:
$$1 - 1 = c + 1 - 1 \Rightarrow$$
$$c = 0$$
Solution 2: A much more obvious solution is to use the fact that \(\lim_{x \rightarrow \infty}\frac{1}{x} = 0\). Rewrite our limit:
$$\lim_{x \rightarrow \infty}\frac{4}{x^2} = 4\lim_{x \rightarrow \infty}\frac{1}{x} \cdot \lim_{x \rightarrow \infty}\frac{1}{x} = 4(0)(0) = 0$$
Example 2: Find the value of the limit \(\lim_{x \rightarrow \infty}\frac{\ln x - x^2}{x^2}\).
Solution: We let this limit equal an arbitrary constant once again:
$$\lim_{x \rightarrow \infty}\frac{\ln x - x^2}{x^2} = c$$
We notice that we can write \(1\) in the form \(\frac{x^2}{x^2}\), so looking to simplify the limit, we add \(1\) to both sides of the equation and apply some algebra to the limit:
$$\lim_{x \rightarrow \infty}{\frac{\ln x - x^2}{x^2}} + 1 = c + 1 \Rightarrow$$
$$\lim_{x \rightarrow \infty}\left(\frac{\ln x - x^2}{x^2} + 1\right) = c + 1 \Rightarrow$$
$$\lim_{x \rightarrow \infty}\left(\frac{\ln x - x^2}{x^2} + \frac{x^2}{x^2}\right) = c + 1 \Rightarrow$$
$$\lim_{x \rightarrow \infty}\left(\frac{\ln x}{x^2}\right) = c + 1$$
The limit is now of the \(\frac{\infty}{\infty}\) form. Thus we can use L'Hopital's Rule on it:
$$\lim_{x \rightarrow \infty}\frac{\frac{1}{x}}{2x} = c + 1 \Rightarrow$$
$$\lim_{x \rightarrow \infty}\frac{1}{2x^2} = c + 1 \Rightarrow$$
$$0 = c + 1 \Rightarrow$$
$$c = -1$$
Usage of The Natural Logarithm
In some even more complicated cases we call upon the natural logarithm for use in evaluating limits. Usually in this case the limit is taking on one of these indeterminate forms:
$$0^0$$
$$\infty^{0}$$
$$\infty^{-\infty}$$
$$(-\infty)^{\infty}$$
$$\infty \cdot 0$$
Just as in the previous section, we let the limit equal some constant, but this time we take the natural logarithm of both sides of this newfound equation to remove any exponents. Oftentimes, when the resulting mess is simplified, L'Hopital's Rule becomes applicable and can be used to finish the problem.
We will start with a famous example often illustrated in calculus classes.
Example 3: Find the value of the limit \(\lim_{x \rightarrow \infty}\left(1 + \frac{1}{x}\right)^x\).
Solution: As before, we set the limit equal to an arbitrary constant:
$$\lim_{x \rightarrow \infty}\left(1 + \frac{1}{x}\right)^x = c$$
Now we take the natural logarithm of both sides of the equation:
$$\ln \lim_{x \rightarrow \infty}\left(1 + \frac{1}{x}\right)^x = \ln c$$
One cool trick to take note of is that the limit notation can be pulled in front of the logarithm (and vice-versa). This is done every time we use the natural logarithm to compute a limit.
$$\lim_{x \rightarrow \infty} \ln \left(1 + \frac{1}{x}\right)^x = \ln c$$
Think of this manipulation as letting the natural logarithm do its work. In this case (and in many others) we use the logarithm identity \(\ln(a^b) = b\ln(a)\) to remove the exponent altogether. Therefore
$$\lim_{x \rightarrow \infty} x \cdot \ln \left(1 + \frac{1}{x}\right) = \ln c$$
So how is L'Hopital's Rule applicable now? We can rewrite the \(x\) outside the natural logarithm as \(\frac{1}{\frac{1}{x}}\).
$$\lim_{x \rightarrow \infty} \frac{\ln \left(1 + \frac{1}{x}\right)}{\frac{1}{x}} = \ln c$$
Now this is a \(\frac{0}{0}\) indeterminate form, so we can use L'Hopital's Rule! Note that to differentiate the numerator we must also use the Chain Rule. It helps to remember that \(\frac{d}{dx}\ln x = \frac{1}{x}\) in this instance.
$$\lim_{x \rightarrow \infty} \frac{-\frac{1}{x^2} \cdot \frac{1}{1 + \frac{1}{x}}}{-\frac{1}{x^2}} = \ln c$$
The \(-\frac{1}{x^2}\) factors cancel, leaving
$$\lim_{x \rightarrow \infty} \frac{1}{1 + \frac{1}{x}} = \ln c$$
Multiply the numerator and the denominator of the fraction by \(x\) to eliminate the complex fraction:
$$\lim_{x \rightarrow \infty} \frac{x}{x + 1} = \ln c$$
This is an \(\frac{\infty}{\infty}\) indeterminate form, so use L'Hopital's Rule again:
$$\lim_{x \rightarrow \infty}\frac{1}{1} = \ln c \Rightarrow$$
$$1 = \ln c$$
One common mistake is to claim the limit equals \(1\) at this point. That is not the case, because we set \(c\) to be the value of the original limit, so we must solve for \(c\). Do this by applying the \(e\) exponent base to both sides of the equation:
$$c = e$$
In summary, \(\lim_{x \rightarrow \infty}\left(1 + \frac{1}{x}\right)^x = e\). Yes, you can memorize the value of this limit, but what is more important are the computations used to get to this fascinating conclusion. Ensure you understand why we can use the natural logarithms the way we did. Read through the proof again if need be. Then, move on to the next set of examples, which explore some of the other indeterminate forms we mentioned.
Example 4: Find the value of the limit \(\lim_{x \rightarrow \infty}(x^2 + e^x)^{-\frac{1}{x}}\).
Solution: This limit is a \(\infty^{0}\) indeterminate form. Therefore we can attempt using natural logarithms. Set that limit to an arbitrary constant:
$$\lim_{x \rightarrow \infty}(x^2 + e^{x})^{-\frac{1}{x}} = c$$
Now take the natural logarithm of both sides of the equation:
$$\ln \lim_{x \rightarrow \infty}(x^2 + e^{x})^{-\frac{1}{x}} = \ln c$$
Invert the natural logarithm and limit:
$$\lim_{x \rightarrow \infty} \ln(x^2 + e^{x})^{-\frac{1}{x}} = \ln c$$
We can now use the natural logarithm to remove the exponent:
$$\lim_{x \rightarrow \infty} -\frac{1}{x}\ln(x^2 + e^{x}) = \ln c$$
Move the negative sign outside of the limit:
$$-\lim_{x \rightarrow \infty} \frac{1}{x}\ln(x^2 + e^{x}) = \ln c$$
Now we can rewrite the limit:
$$-\lim_{x \rightarrow \infty}\frac{\ln(x^2 + e^x)}{x} = \ln c$$
Now this limit is of the \(\frac{\infty}{\infty}\) indeterminate form, so we can use L'Hopital's Rule:
$$-\lim_{x \rightarrow \infty}\frac{2x + e^x}{x^2 + e^x} = \ln c$$
This limit is still of the \(\frac{\infty}{\infty}\) indeterminate form, so we use L'Hopital's Rule again:
$$-\lim_{x \rightarrow \infty}\frac{2 + e^x}{2x + e^x} = \ln c$$
It is still of the same indeterminate form, so repeat L'Hopital's Rule:
$$-\lim_{x \rightarrow \infty}\frac{e^x}{2 + e^x} = \ln c$$
Using L'Hopital's Rule one last time gives
$$-\lim_{x \rightarrow \infty}\frac{e^x}{e^x} = \ln c \Rightarrow$$
$$-1 = \ln c \Rightarrow$$
$$c = \frac{1}{e}$$