Find slope of tangent line using m$_{tan}=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$ and the point $P = (5,\frac{2}{5})$

Question

Perhaps I'm missing something simple here, but every time I attempt this problem I get the same answer that does not make sense.

The question says, use the definition m$_{tan}=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$ to find the slope of the line tangent to the graph of $f$ at $P$. I am given $f(x)=\frac{2}{x}$ and $P = (5,\frac{2}{5})$

My work:

let $a=5$:

m$_{tan}=\lim\limits_{x\to 5}\dfrac{\dfrac{2}{x}-f(5)}{x-5}$

m$_{tan}=\lim\limits_{x\to 5}\dfrac{\dfrac{2}{x}-\dfrac{2}{5}}{x-5}$

Simplify:

m$_{tan}=\lim\limits_{x\to 5}\dfrac{\dfrac{2(5-x)}{5x}}{x-5}$

m$_{tan}=\lim\limits_{x\to 5}\dfrac{2(5-x)}{5x}\cdot\dfrac{1}{x-5}$

m$_{tan}=\lim\limits_{x\to 5}\dfrac{2(5-x)}{5x(x-5)}$

Now when you plug in the limit you get $\frac{0}{0}$ which makes no sense.

Can anyone point out where I'm making the mistake?

I was about to say they are not the same, but I just realized you can just pull a negative out...wow, thank you! If you want to change to answer I will accept. — Sabien, Aug 29 '14 at 00:55
For a discussion of the issues regarding "plugging in $x=5$" you might like to read my answer to this question. — David, Aug 29 '14 at 00:59

score 0 · Accepted Answer · answered Aug 29 '14 at 01:21

0

Notice that $lim \frac{2(5-x)}{5x} . \frac{1}{x-5} = lim \frac{-2(x-5)}{5x} . \frac{1}{(x-5)} = lim \frac{-2}{5x} = \frac{-2}{25}$ as $x \rightarrow 5$

answered Aug 29 '14 at 01:21

Aaron Maroja

17,571

Find slope of tangent line using m$_{tan}=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$ and the point $P = (5,\frac{2}{5})$

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