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.

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This entry was posted on Monday, September 13th, 2010 at 12:17 am and is filed under 170: Calculus I. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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