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I don't know if "identity" is the correct word, but this would be an example: $$\lim_{x \to \infty} \left(1 + \frac{1}{x} \right)^{x} = e $$

With my current knowledge, I wouldn't really know where to begin in solving this, so whenever I see this notation (or something similar like $\lim_{x \to \infty} \left(1 - \frac{1}{x} \right)^{x} = \frac{1}{e} $), I know I'm dealing with some form of $e$ (if that makes sense).

Other properties I have memorized are for example what the limit of $\arctan$ is (even though I could simply visualize it, we can't using graphing calculators on the exam and I'm very bad at sketching more complex functions), or that $\lim_{x \to 0} \frac{a}{x} \sin{x} = a$. I also just learned about the Stirling approximation from another user here.

Are there any other noticeable limits like these that I should be on lookout for?

Community
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Andres Stadelmann
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3 Answers3

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Hint

From the other user !!

$$A= \left(1 + \frac{1}{x} \right)^{x}$$ $$\log(A)=x\log\left(1 + \frac{1}{x} \right)$$ Now, remember that when $y$ is small $\log(1+y)\approx y$ Replace $y$ by $\frac 1x$ and you are done.

What you must know, remember and use are Taylor series for the basic functions.

Claude Leibovici
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    especially for $e^x and \sin(x)$ !!!! – Airdish Jan 07 '16 at 14:47
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    @S.Mo. Plus many other ! I started being in love with Taylor series almost 60 years ago. Back to serious, it is more important to know how to build them from scratch (which is quite easy). But memorizing some of them is really crucial. – Claude Leibovici Jan 07 '16 at 14:57
  • I need to learn how to make Taylor series by myself since my professor failed to finish the curriculum (I feel stupid only telling myself this now). From what I understand it is an approximation of a function by adding other simpler functions (kind of what the Fourier series does with trig functions - why I learned this before, don't ask me), but how does this help to solve limits? – Andres Stadelmann Jan 07 '16 at 15:31
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    @AndresStadelmann. It is just a polynomial expansion built around a value. Then, you baiscally have to consider how behaves the ratio of two polynomials close to a point. Then, the limit is simple. Look in this site; there are thousands of applications. – Claude Leibovici Jan 07 '16 at 15:41
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    @AndresStadelmann Taylor series helps you to solve limits because it allows you to sort of play with a function when its limit is otherwise indeterminate. – Airdish Jan 07 '16 at 17:56
  • It took me a day, but I realize now that you were giving me a hint for how to solve $\left(1 + \frac{1}{x} \right)^{x}$. Thanks so much! I have another question regarding Taylor series (which are indeed pretty easy to build), can they be used to approximate the infinite sum of $\frac{1}{x^{2}}$ for example? Or are they mostly just useful to solve limits... – Andres Stadelmann Jan 08 '16 at 17:29
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Note: this is not an answer. I just didn't want to deal with writing latex in the comments.

A few off the top of my head are $\lim_{x\to 0} {\dfrac{sinx}{x}} = 1$ and $\lim_{x \to 0}{\dfrac{e^x - 1}{x} = 1}$

Also $\lim_{x \to a}{(1+\dfrac{1}{f(x)})^{f(x)} = e}$ and $\lim_{x \to a}({\dfrac{f(x)}{g(x)}})^{h(x)} = e^{\lim_{x \to a}{h(x)(\dfrac{f(x)}{g(x)} - 1)}}$

Airdish
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  • I've seen the proof of why $\lim_{x\to 0} {\dfrac{sinx}{x}} = 1$ and subsequently $\lim_{x\to \infty} \frac{sin{\frac{1}{x}}}{\frac{1}{x}} = 1$. I've always wondered how $\lim_{x\to \infty} \frac{sin{\frac{a}{x}}}{\frac{1}{x}} = a$ is equivalent to $\lim_{x\to 0} {\dfrac{sinx}{\frac{x}{a}}} = a$... – Andres Stadelmann Jan 07 '16 at 15:21
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If you will be encountering sequences and series this is an important one to know:

$$\sum_\limits{n=1}^{\infty} \frac{1}{n} =\lim \limits_{n \to \infty} (1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}) \rightarrow \infty$$

This is the harmonic sequence, and is very important in determining the convergence of infinite series. The limit can be shown easily using the integral test.

Edit: I also think that this is particularly important:

$$\lim \limits_{n \to \infty} \frac{A}{n}=0$$ Where $A$ is any real number. Intuitively, this means that an finite number "divided by infinity" will be nothing. That is, a finite quantity, no matter how large, divided by an infinite "quantity" is zero.

zz20s
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