Limit question / algebra problem

[tony873004]

I forgot how to do the algebra in this one.

$$\begin{array}{l} f(x) = \frac{{ - 1}}{x};\,\,a = 1 \\ \\ f'(a) = \mathop {\lim }\limits_{h \to 0} \frac{{f(a + h) - f(a)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - 1}}{{a + h}} - \frac{{ - 1}}{a}}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - 1a}}{{a(a + h)}} - \frac{{ - 1(a + h)}}{{a(a + h)}}}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - a - a + h}}{{a(a + h)}}} \right)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - 2a + h}}{{a^2 + ah}}} \right)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{ - 2a + h}}{{a^2 + ah}}\,\,\frac{1}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{ - 2a + h}}{{h\left( {a^2 + ah} \right)}} \\ \end{array}$$

Am I even going in the right direction? If so, what comes next?

 

[d_leet]

You screwed up adding the fractions. -(-1) = 1.

 

[tony873004]

That's a bit nicer. Thanks d leet
$$\begin{array}{l} f(x) = \frac{{ - 1}}{x};\,\,a = 1 \\ \\ f'(a) = \mathop {\lim }\limits_{h \to 0} \frac{{f(a + h) - f(a)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - 1}}{{a + h}} - \frac{{ - 1}}{a}}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - 1a}}{{a(a + h)}} - \frac{{ - 1(a + h)}}{{a(a + h)}}}}{a} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - a - \left( { - a - h} \right)}}{{a(a + h)}}} \right)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - a + a + h}}{{a(a + h)}}} \right)}}{h} \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - a + a + h}}{{a(a + h)}}} \right)}}{h} \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{h}{{a^2 + ah}}} \right)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{h}{{a^2 + ah}}\,\,\frac{1}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{h}{{h\left( {a^2 + ah} \right)}} = \\ \\ f'(a) = \frac{1}{{a^2 }} \\ \end{array}$$
Now I can't find my TEX error, but I get the right answer: 1/a^2

 

[VietDao29]

$$\begin{array}{l} f(x) = \frac{{ - 1}}{x};\,\,a = 1 \\ \\ f'(a) = \mathop {\lim }\limits_{h \to 0} \frac{{f(a + h) - f(a)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - 1}}{{a + h}} - \frac{{ - 1}}{a}}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - 1a}}{{a(a + h)}} - \frac{{ - 1(a + h)}}{{a(a + h)}}}}{a} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - a - \left( { - a - h} \right)}}{{a(a + h)}}} \right)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - a + a + h}}{{a(a + h)}}} \right)}}{h} \\ \end{array}$$
$$\begin{array}{l} \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{{ - a + a + h}}{{a(a + h)}}} \right)}}{h} \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{{\left( {\frac{h}{{a^2 + ah}}} \right)}}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{h}{{a^2 + ah}}\,\,\frac{1}{h} = \\ \\ \mathop {\lim }\limits_{h \to 0} \frac{h}{{h\left( {a^2 + ah} \right)}} = \\ \\ f'(a) = \frac{1}{{a^2 }} \\ \end{array}$$
I think you should split it up a bit, it seems to work if you split it into 2 LaTeX images. However, IMHO, you are using way too many brackets of the type {...}. It should be clearer if you you less of them. Say, instead of typing \frac{{-1}}{a}, just type:
\frac{-1}{a}, which also displays the correct image, but the code does look nicer, right?
$$\frac{-1}{a}$$
Don't complicate things... :)

 

[tony873004]

Thanks, VietDao. I was having some trouble with the forum. Clicking on my code wouldn't open it so I couldn't edit it.

Actually, I didn't hand-code this TEX. I use Math-type, which translates it into tex for me. In a sense it's like using MS Word to write a simple web page. Hand-coding HTML is short and to the point, and the code looks nice. But the Word version, even in the simplest example is mind-numbingly complex.

I'm better with Math-type than tex, and Math-type let's you see what you're typing, i.e., WYSIWYG. So for the longer problems I use Math-type and have it translate.

Thanks for fixing it for me. It was frusterating me more than dumb mistake than d leet pointed out to me.