Answer
There are theorems to calculate limits, which are detailed in the Work step by step below.
Work Step by Step
There are 5 theorems to calculate limits.
1) Theorem 1: Limit Laws
Suppose that $\lim_{x\to c}f(x)=L$ and $\lim_{x\to c}g(x)=M$
- Sum Rule: $\lim_{x\to c}(f(x)+g(x))=L+M$
$\lim_{x\to c}(2+3x)=\lim_{x\to c}2+\lim_{x\to c}3x$
- Difference Rule: $\lim_{x\to c}(f(x)-g(x))=L-M$
$\lim_{x\to c}(2-3x)=\lim_{x\to c}2-\lim_{x\to c}3x$
- Constant Multiple Rule: $\lim_{x\to c}(kf(x))=kL$
$\lim_{x\to c}3x=3\lim_{x\to c}x$
- Product Rule: $\lim_{x\to c}(f(x)g(x))=LM$
$\lim_{x\to c}(6x)=\lim_{x\to c}6\times\lim_{x\to c}x$
- Quotient Rule: $\lim_{x\to c}\frac{f(x)}{g(x)}=\frac{L}{M}$ $(M\ne0)$
$\lim_{x\to c}\frac{x^2-1}{x+2}=\frac{\lim_{x\to c}(x^2-1)}{\lim_{x\to c}(x+2)}$
- Power Rule: $\lim_{x\to c}[f(x)]^n=L^n$ ($n$ is a positive integer)
$\lim_{x\to c}16x^2=(\lim_{x\to c}4x)^2$
- Root Rule: $\lim_{x\to c}\sqrt[n]{f(x)}=\sqrt[n]L$ ($n$ is a positive integer and if $n$ is even, we assume $L\gt0$)
$\lim_{x\to c}\sqrt[3]{4x^2}=\sqrt[3]{\lim_{x\to c}4x^2}$
2) Theorem 2: Limits of Polynomials
If $P(x)=a_nx^{n}+a_{n-1}x^{n-1}+...+a_0$ then $\lim_{x\to c}P(x)=P(c)=a_nc^n+a_{n-1}c^{n-1}+...+a_0$
- Take this example: $$\lim_{x\to2}(x^3-2x^2+5x+1)=2^3-2\times2^2+5\times2+1=11$$
3) Theorem 3: Limits of Rational Functions
If $P(x)$ and $Q(x)$ are polynomials and $Q(c)\ne0$, then $\lim_{x\to c}\frac{P(x)}{Q(x)}=\frac{P(c)}{Q(c)}$
$$\lim_{x\to 2}\frac{x^3+7}{x^2-1}=\frac{2^3+7}{2^2-1}=\frac{15}{3}=5$$
4) Theorem 4: Sandwich Theorem
Suppose $f(x)\le g(x)\le h(x)$ for all $x$ in an open interval containing $c$, but not necessarily at $x=c$. And suppose that $\lim_{x\to c}f(x)=\lim_{x\to c}h(x)=L$. Then $\lim_{x\to c}g(x)=L$
- Example: Find $\lim_{x\to 0}x\sin x$
Since $-x\le x\sin x\le x$ and $\lim_{x\to 0}-x=\lim_{x\to 0}x=0$, we conclude that $\lim_{x\to 0}x\sin x=0$ according to Sandwich Theorem.
5) Theorem 5: If $f(x)\le g(x)$ for all $x$ in some open interval containing $c$, but not necessarily at $c$, and $\lim_{x\to c}f(x)$ and $\lim_{x\to c}g(x)$ both exist, then $\lim_{x\to c}f(x)\le\lim_{x\to c}g(x)$
- Example: We know that $x\le|x|$ for all $x$
So $\lim_{x\to c}x\le\lim_{x\to c}|x|$ for all $x$ as well.
