# Statement of the limit laws and first examples

• Suppose that $$c$$ is a constant, and the limits $$\lim_{x\to a}f(x)$$ and $$\lim_{x\to a}g(x)$$ exist. Then we have the following formulas.
• (The reason for the strange numbering is that I wanted them to agree with the numbering of the textbook.)
1-2. Sum and Difference law. $$\lim_{x\to a}(f(x)\pm g(x))=(\lim_{x\to a}f(x))\pm(\lim_{x\to a}g(x))$$
1. Product law. $$\lim_{x\to a}f(x)g(x)=(\lim_{x\to a}f(x))(\lim_{x\to a}g(x))$$
2. Quotient law. If $$\lim_{x\to a}g(x)\ne0$$, then we have $$\lim_{x\to a}\frac{f(x)}{g(x)}=\frac{\lim_{x\to a}f(x)}{\lim_{x\to a}g(x)}$$.
3. $$\lim_{x\to a}c=c$$
4. $$\lim_{x\to a}x=a$$.
5. Constant multiple law. Rules 4 and 7 give $$\lim_{x\to a}(cf(x))=c\lim_{x\to a}f(x)$$.
6. Power law. Let $$n$$ be a positive integer. Applying the Product law repeatedly for $$g(x)=f(x)$$, we get $$\lim_{x\to a}(f(x))^n=(\lim_{x\to a}f(x))^a$$.
7. Root law. Let $$n$$ be a positive integer. In case $$n$$ is even, let's assume that $$\lim_{x\to a}f(x)>0$$. Then we have $$\lim_{x\to a}\sqrt[n]{f(x)}=\sqrt[n]\lim_{x\to a}f(x)$$.
• Exercises. 2.3: 4,8
• Direct substitution property. If $$f(x)$$ is a polynomial or a rational function, and $$a$$ is in the domain of $$f(x)$$, then we have $$\lim_{x\to a}f(x)=f(a)$$.
• All the limit laws have one-sided versions.

# More examples

• Recall that we have guessed $$\lim_{x\to1}\frac{x-1}{x^2-1}=\frac12$$.
• Note that we can't use direct substitution because $$f(x)=\frac{x-1}{x^2-1}$$ is not defined at $$x=1$$
• We can't use the Quotient law either, because the denominator $$x-1$$ has limit $$0$$ as $$x\to1$$.
• But we can use that when calculating this limit, we only need to consider values $$x$$ near $$1$$ and not equal to $$1$$.
• Therefore, from the fraction $$\frac{x-1}{x^2-1}=\frac{x-1}{(x-1)(x+1)}$$ we can cancel the common factor $$(x-1)$$ inside the limit: $\lim_{x\to1}\frac{x-1}{x^2-1}=\lim_{x\to1}\frac{x-1}{(x-1)(x+1)}=\lim_{x\to1}\frac{1}{x+1}=\frac12.$
• The same rule holds in general: Suppose that the limits $$\lim_{x\to a}f(x)$$ and $$\lim_{x\to a}g(x)$$ exist, and that we have $$f(x)=g(x)$$ when $$x\ne a$$. Then we have $$\lim_{x\to a}f(x)=\lim_{x\to a}g(x)$$.
• Let $g(x)=\begin{cases} x+1& x\ne1,\\ \pi & x=1. \end{cases}$
• Since $$g(x)=x+1$$ whenever $$x\ne1$$, we have $\lim_{x\to1}g(x)=\lim_{x\to1}(x+1)=2.$
• Exercises. 2.3: 12,20,28,32.

# $$\lim_{x\to0}\frac{\sqrt{t^2+9}-3}{t^2}$$

• We can now calculate this limit, which was trouble when we tried to do this numberically.
• The trick is to get rid of the square root in the numerator: $\frac{\sqrt{t^2+9}-3}{t^2}=\frac{\sqrt{t^2+9}-3}{t^2}\frac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3}=\frac{t^2+9-9}{t^2(\sqrt{t^2+9}+3)}=\frac{1}{\sqrt{t^2+9}+3}.$
• With this, we can calculate the limit: $\lim_{t\to0}\frac{\sqrt{t^2+9}-3}{t^2}=\lim_{t\to0}\frac{1}{\sqrt{t^2+9}+3}=\frac16.$
• Exercises. 2.3: 22,30,34,56,61-63