Limits by Logarithms, Squeeze Theorem, and Euler’s Constant
Various tricks for evaluating tricky limits.
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Limits by Logarithms, Squeeze Theorem, and Euler's Constant. In Justin Math: Calculus. https://justinmath.com/limits-by-logarithms-squeeze-theorem-and-eulers-constant/
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A useful property of limits is that they can be brought inside continuous functions, i.e. the limit of a continuous function is the function of the limit.
For example, $\sqrt{x}$ is a continuous function, so to take the limit of the square root of some expression, we can first find the limit of the expression and then take the square root.
We can do the same thing with other continuous functions, such as $\ln x$.
Exponential Limits
Logarithms in particular are useful for evaluating exponential limits, which have variables in both the limit and the base.
For example, to evaluate the limit
it is easiest to start by evaluating the logarithm of the limit.
Since we know the logarithm of the limit is $\infty$, the limit is just $e$ raised to the power of $\infty$.
Using the same process, we can show that
because this time, the logarithm of the limit evaluates to $-\infty$.
Squeeze Theorem
Another useful trick for evaluating difficult limits is squeezing them between limits that are easier to evaluate.
For example, to evaluate the limit
we can make use of the fact that $\sin x$ is bounded between $-1$ and $1$. Then as $x \to \infty$ we have the following:
The inequality states that the limit must be between $0$ and $0$, and the only number that is between $0$ and $0$ is $0$ itself, so by the squeeze theorem, the limit must evaluate to $0$.
In other words, the limit must be $0$ because we squeezed it between two other limits, both of which evaluate to $0$.
As another example, we can show that
by performing a squeeze between the bounds of $\cos$:
Euler's Constant
Lastly, Euler’s constant $e$ can be expressed as the following limit:
It also holds as $n \to -\infty$:
Substituting $x=\frac{1}{n}$, we can also express the limit as
Knowing the above limit forms of Euler’s constant allows us to compute limits that are in a similar form. For example, to compute the limit
we can make a substitution that results in $\frac{2}{n} = \frac{1}{u}$. Then $n=2u$, and $n \to \infty$ translates to $u \to \infty$, and the limit becomes computable in terms of Euler’s constant:
Similarly, to compute the limit
we can make a substitution that results in $-3x=u$. Then $x=-\frac{1}{3}u$, and $x \to 0$ translates to $u \to 0$, and the limit becomes computable in terms of Euler’s constant:
Exercises
Evaluate the following limits using logarithms. (You can view the solution by clicking on the problem.)
$\begin{align*}1) \hspace{.5cm} \lim\limits_{x\to 0^+} x^{(\ln x)^3} \end{align*}$
Solution:
$\infty$
$\begin{align*}2) \hspace{.5cm} \lim\limits_{x\to 0^+} x^{(\ln x)^4} \end{align*}$
Solution:
$0$
$\begin{align*}3) \hspace{.5cm} \lim\limits_{x\to 0^+} x^{\frac{1}{\ln x}} \end{align*}$
Solution:
$e$
$\begin{align*}4) \hspace{.5cm} \lim\limits_{x\to 0^+} x^{\frac{1}{(\ln x)^2}} \end{align*}$
Solution:
$1$
Evaluate the following limits using the squeeze theorem. (You can view the solution by clicking on the problem.)
$\begin{align*}5) \hspace{.5cm} \lim\limits_{x\to \infty} \frac{\sin x + \cos x}{\ln x} \end{align*}$
Solution:
$0$
$\begin{align*}6) \hspace{.5cm} \lim\limits_{x\to 0^+} \sqrt{x} \cos \left( \frac{1}{1-e^x} \right) \end{align*}$
Solution:
$0$
$\begin{align*}7) \hspace{.5cm} \lim\limits_{x\to \infty} \frac{3x+\sin x}{\sqrt{3x^2-1}} \end{align*}$
Solution:
$\sqrt{3}$
$\begin{align*}8)\hspace{.5cm} \lim\limits_{x\to 1^+} (\ln x) \sin \left( \frac{1}{\ln x} \right) \end{align*}$
Solution:
$0$
$\begin{align*}9) \hspace{.5cm} \lim\limits_{x\to -\infty} \frac{4x+3\cos x}{2x+1} \end{align*}$
Solution:
$2$
$\begin{align*}10)\hspace{.5cm} \lim\limits_{x\to \infty} \frac{2x+x\sin x}{\sqrt{x^3-4}} \end{align*}$
Solution:
$0$
Evaluate the following limits using Euler’s constant. (You can view the solution by clicking on the problem.)
$\begin{align*}11) \hspace{.5cm} \lim\limits_{n \to \infty} \left( 1 + \frac{1}{2n} \right)^{4n} \end{align*}$
Solution:
$e^2$
$\begin{align*}12) \hspace{.5cm} \lim\limits_{x \to 0} \left( 1 + \frac{2}{3}x \right)^{\frac{1}{2x}}\end{align*}$
Solution:
$e^\frac{1}{3}$
$\begin{align*}13) \hspace{.5cm} \lim\limits_{n \to -\infty} \left( 1 + \frac{5}{3n} \right)^{-2n}\end{align*}$
Solution:
$e^{ - \frac{10}{3} }$
$\begin{align*}14) \hspace{.5cm} \lim\limits_{x \to 0} \left( 1 + \frac{x}{\pi} \right)^{\frac{e}{x}}\end{align*}$
Solution:
$e^\frac{e}{\pi}$
This post is part of the book Justin Math: Calculus. Suggested citation: Skycak, J. (2019). Limits by Logarithms, Squeeze Theorem, and Euler's Constant. In Justin Math: Calculus. https://justinmath.com/limits-by-logarithms-squeeze-theorem-and-eulers-constant/
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