Using formal definition of a limit

Mar 2017
358
3
Massachusetts
Can someone help me complete the derivative computation below? I'm a bit stuck.

IMG_20190615_161557.jpg
 
Last edited:

Walagaster

MHF Helper
Apr 2018
230
146
Tempe, AZ
Can someone help me complete the derivative computation below? I'm a bit stuck.

View attachment 39421
Your last 3 steps should not have the limit statement in front of them because you already took the limit. The next to last step to the last step is wrong:
$\frac{4-2a}{2-a} = \frac{2(2-a)}{2-a} = 2$
It would be much better for you to type the equations so we could edit your post.
 
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Dec 2014
166
126
USA
$g(x)=2x-1$

derivative of $g$ at any value $x=a$ in the domain of $g$

$\displaystyle g’(a) = \lim_{x \to a} \dfrac{g(x)-g(a)}{x-a}$

$\displaystyle g’(a) = \lim_{x \to a} \dfrac{2x-1 - (2a-1)}{x-a}$

$\displaystyle g’(a) = \lim_{x \to a} \dfrac{2x-2a}{x-a}$

$\displaystyle g’(a) = \lim_{x \to a} \dfrac{2(x-a)}{x-a}$


$\displaystyle g’(a) = \lim_{x \to a} \dfrac{2( \cancel {x-a})}{\cancel {x-a}} = 2$
 
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