Hello guys i have a homework question that once again has me stumped. I have a less then friendly teacher who thinks all her students are going to be the next Isaac Newton and she barely explains anything.
So here is my question.
from reading my book it appears that i have two options.
formula 1 says
$$ \lim_{x\to a }\frac{f(x)-f(a)}{x-a}$$
and the second says
$$ \lim_{h\to 0 }\frac{f(a+h)-f(a)}{h}$$
so i tried to use formula 2 to solve this problem
FIND The equation of the tangent line to the curve at the given point.
$$ \sqrt{x},$$ $$ (1,1)$$
step – 1
$$ \frac{f(1+h)-f(1)}{h}$$
step -2
$$\frac{\sqrt{1+h} – 1}{h}$$
step -3 used the conjugate pair to come up with
$$\frac{\sqrt{1+h} – 1}{h} * \frac{\sqrt{1+h} + 1}{\sqrt{1+h} + 1} $$
step -4
the result of the above i would end up with
$$ \frac{1+h-1}{h\sqrt{1+h}}$$
step 5
$$\frac{1}{\sqrt{1+0}} = 1$$
however the answer is $ \frac{1}{2}$ i cant see what I'm doing wrong and when should i use the first formula and when do i use the second formula? any help in answering this question would be most appreciated.
Thanks
Miguel
Best Answer
You lost a one going from step 3 to step 4. More exactly:
$$ \frac{\sqrt{1+h} - 1}{h} \frac{\sqrt{1+h} + 1}{\sqrt{1+h} {\bf+ 1}}=\frac{1+h - 1}{h(\sqrt{1+h} {\bf+ 1})}$$