## 170- Quiz 3

Quiz 3 is here.

Solutions follow.

Problem 1 asks to find $\displaystyle \lim_{x\to 0^+}\frac x{|x|}$, $\displaystyle \lim_{x\to 0^-}\frac x{|x|}$, and $\displaystyle \lim_{x\to 0}\frac x{|x|}$.

Recall that $x\to 0^+$ means we are to consider values of $x$ that are closer and closer to 0, and positive, while $x\to 0^-$ means we consider values of $x$ that are closer and closer to 0, and negative.

Note that if $x>0$, then ${}|x|=x$, so $\displaystyle \frac x{|x|}=\frac {x}{x}=1$. This expression is a constant, so its value does not change as $x\to 0^+$, and we have

$\displaystyle \lim_{x\to0^+}\frac x{|x|}=1.$

Similarly, if $x<0$, then ${}|x|=-x$; think, for example, of $x=-3$. Then $|x|=3=-(-3)$. It follows that $\displaystyle \frac x{|x|}=\frac {x}{-x}=-1$. This expression is a constant, so its value does not change as $x\to 0^-$, and we have

$\displaystyle \lim_{x\to0^-}\frac x{|x|}=-1.$

Finally, since $\displaystyle \lim_{x\to0^+}\frac x{|x|}\ne \lim_{x\to0^-}\frac x{|x|}$, it follows that

$\displaystyle \lim_{x\to0}\frac x{|x|}$ does not exist.

Problem 2 asks to find a formula, as simple as possible, for $f'(x)$ when $f(x)=100-10x^2$.

We have that $\displaystyle f'(x)=\lim_{\triangle x\to 0}\frac{f(x+\triangle x)-f(x)}{\triangle x}$. Since $f(x)=100-10x^2$, we have that $f(x+\triangle x)=100-10(x+\triangle x)^2$ $=100-10(x^2+2x\triangle x+(\triangle x)^2)$ $=100-10x^2-20x\triangle x-10(\triangle x)^2$ and $f(x+\triangle x)-f(x)$ $=100-10x^2-20x\triangle x-10(\triangle x)^2-(100-10x^2)$ $=-20x\triangle x-10(\triangle x)^2$, so

$\displaystyle \frac{f(x+\triangle x)-f(x)}{\triangle x}=\frac{-20x\triangle x-10(\triangle x)^2}{\triangle x}=-20 x-10\triangle x,$

and as $\triangle x\to 0$, this expression approaches $-20x-0=-20x$, i.e., $f'(x)=-20x$.