2.3 The limit laws  (Page 4/15)

[link] illustrates the function $f\left(x\right)$ and aids in our understanding of these limits.

1. Since $f\left(x\right)=4x-3$ for all x in $\left(\text{−}\infty ,2\right),$ replace $f\left(x\right)$ in the limit with $4x-3$ and apply the limit laws:
   
   $\underset{x\to 2^{-}}{\text{lim}}f\left(x\right)=\underset{x\to 2^{-}}{\text{lim}}\left(4x-3\right)=5.$
2. Since $f\left(x\right)={\left(x-3\right)}^2$ for all x in $\left(2,\text{+}\infty \right),$ replace $f\left(x\right)$ in the limit with ${\left(x-3\right)}^2$ and apply the limit laws:
   
   $\underset{x\to 2^{+}}{\text{lim}}f\left(x\right)=\underset{x\to 2^{-}}{\text{lim}}{\left(x-3\right)}^2=1.$
3. Since $\underset{x\to 2^{-}}{\text{lim}}f\left(x\right)=5$ and $\underset{x\to 2^{+}}{\text{lim}}f\left(x\right)=1,$ we conclude that $\underset{x\to 2}{\text{lim}}f\left(x\right)$ does not exist.

Step 1. After substituting in $x=2,$ we see that this limit has the form $\mathrm{-1}\text{/}0.$ That is, as x approaches 2 from the left, the numerator approaches −1; and the denominator approaches 0. Consequently, the magnitude of $\frac{x-3}{x(x-2)}$ becomes infinite. To get a better idea of what the limit is, we need to factor the denominator:

Step 2. Since $x-2$ is the only part of the denominator that is zero when 2 is substituted, we then separate $1\text{/}(x-2)$ from the rest of the function:

Step 3. $\underset{x\to 2^{-}}{\text{lim}}\frac{x-3}{x}=-\frac{1}{2}$ and $\underset{x\to 2^{-}}{\text{lim}}\frac{1}{x-2}=\text{−}\infty .$ Therefore, the product of $(x-3)\text{/}x$ and $1\text{/}(x-2)$ has a limit of $\text{+∞:}$

Because $\mathrm{-1}\le \text{cos}x\le 1$ for all x , we have $-x\le x\text{cos}x\le x$ for $x\ge 0$ and $-x\ge xcosx\ge x$ for $x\le 0$ (if x is negative the direction of the inequalities changes when we multiply). Since $\underset{x\to 0}{\text{lim}}\left(-x\right)=0=\underset{x\to 0}{\text{lim}}x,$ from the squeeze theorem, we obtain $\underset{x\to 0}{\text{lim}}x\text{cos}x=0.$ The graphs of $f\left(x\right)=-x,g\left(x\right)=x\text{cos}x,$ and $h\left(x\right)=x$ are shown in [link] .