Calculation of Limits

Remark 3.2.1 (Convergence of Sequences)  
The definition of continuity greatly increases our calculation power. Because, if we know a function to be continuous, we can use it in the calculation of limits.
An obvious example is

$$\lim_{n \to \infty}\sin\left(\frac{\pi}{n}\right) = \sin\left(\lim_{n \to \infty}\frac{\pi}{n}\right) = \sin (0) = 0$$ (3.2.1)

where the continuity of the $\sin$ function allows us to bring the limit inside.
A more involved example is the calculation of

$$\lim_{n \to \infty}\frac{e^{\sqrt{n}}}{n^n}$$ (3.2.2)

Using the fact that $\ln$ is a continuous function over $\hbox {I \hskip -5.2pt {R}}^+_0$ yields the following result:

$$\ln\left(\lim_{n \to \infty}\frac{e^{\sqrt{n}}}{n^n}\right) = \li... ...c{e^{\sqrt{n}}}{n^n}\right) = \lim_{n \to \infty}\sqrt{n} - n \ln (n) = -\infty$$ (3.2.3)

Finally, if $\ln$ of a sequence goes to $-\infty$, then this sequence goes to 0. Therefore

$$\lim_{n \to \infty}\frac{e^{\sqrt{n}}}{n^n} = 0$$ (3.2.4)

Examples 3.2.2 (Limits for $x \to c$)  
With the long list of continuous functions at our disposal, we now have the ability to calculate about every limit that crosses our path. Look e.g. at

$$\lim_{x \to 3} \frac{x^2+1}{x+1}$$ (3.2.5)

As this function is continuous at $3$ (or at every point except $-1$ for that matter), we can take the limit inside and we obtain:

$$\lim_{x \to 3} \frac{x^2+1}{x+1} = \lim_{x \to 3} \frac{10}{4} = \frac{5}{2}$$ (3.2.6)

Still, there are a couple standard tricks around that have no grand theoretical basis but are used over and over again.
In rational functions, one often encounters $\frac{0}{0}$, which is undefined. Often, eliminating common factors in numerator and denominator solve the problem:

$$\lim_{x \to 1} \frac{x^2-2x+1}{x^2-1} = \lim_{x \to 1} \frac{(x-1)^2}{(x-1)(x+1)} \cr$$ (3.2.7)

Sometimes (especially when roots are involved), it might not be so clear that there is a common factor. In this case, try to multiply numerator and denominator with a conjugated term, e.g. :

$$\lim_{x \to 4} \frac{\sqrt{x}-2}{x-4} = \lim_{x \to 4} \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)} \cr$$ (3.2.8)

When dealing with limits of trigonometric functions, it often simply pays off to know a bunch of trig identities:

$$\lim_{x \to 0} \frac{\sin(2x)}{\sin(x)} = \lim_{x \to 0} \frac{2\sin(x)\cos(x)}{\sin(x)} = 2 \lim_{x \to 0} \cos(x) = 2$$ (3.2.9)

A handy formula to remember is

$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$ (3.2.10)

This limit is easy to prove using de l'Hôpital's rule (4.2.7), but the elementary geometric proof is a real pain in the neck, so we'll skip it for now and forever. The corresponding identity for the cosine function is easily proven using (3.2.10):

$$\lim_{x \to 0} \frac{\cos(x)-1}{x} = \lim_{x \to 0} \frac{\cos^2(x)-1}{x(\cos(x)+1)} \cr$$ (3.2.11)

Examples 3.2.3 (Limits for $x \to \pm\infty$)  
The tricks that apply for limits going to a point also apply for limits going to infinity, yet there are some peculiarities that are almost unavoidable. The most common one is dividing the numerator and the denominator in a rational function by the highest order term:

$$\lim_{x \to +\infty} \frac{6x^3+7x+8}{2x^3+1} = \lim_{x \to +\infty} \frac{6+\frac{7}{x^2}+\frac{8}{x^3}}{2+\frac{1}{x^3}} \cr$$ (3.2.12)

Note that the limit equals the coefficient of the highest order term in the numerator divided by the coefficient of the highest order term in the denominator. Now, do not memorise this rule. If you start going down that road, you will have a lot of similar looking rules to memorise and your brain, being as unreliable as a human brain invariably is, will start mixing them up. As long as you understand the above calculation, you can reproduce it easily in any similar situation.
Another typical problem is calculating a limit like

$$\lim_{x \to +\infty} e^{-x} \sin(x)$$ (3.2.13)

The problem is that although $e^{-x}$ tails of to 0 quit fast, the limit cannot be calculated by simply taking the product of the $2$ limits as the limit for $x$ going to infinity of the sine function is undefined. A way around is is to invoke the squeeze lemma (2.0.17) and notice the following:

$$0 \leq \lim_{x \to +\infty} \left\vert e^{-x} \sin(x) \right\vert \leq \lim_{x \to +\infty} e^{-x} = 0$$ (3.2.14)

So $\lim_{x \to +\infty} e^{-x} \sin(x) = 0$. A little more involved is the calculation of the following limit:

$$\lim_{x \to +\infty} e^{-x} x^2$$ (3.2.15)

It can be calculated by noticing that $\ln(x) \leq \frac{x}{2}$ for any $x$, in fact for large $x$ the difference is quite substantial. Then we have

$$\lim_{x \to +\infty} e^{-x} x^2 = \lim_{x \to +\infty} e^{\ln(x^2)-x} \cr$$ (3.2.16)

This limit will also turn out to be an easy prey for de l'Hôpital's rule (4.2.7).