Calculus Limit: limit, x-->a, of [cos(x) - cos(a)] / (x - a)
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Re: Calculus Limits
Hello DJO:
Please post some clues as to what you already know about these types of exercises. Show any work and reasoning, if you can.
If you already understand the question, then are you able to say something about why you're stuck?
(Please read the post titled, "Read Before Posting", if you have not already done so. We do not, in general, provide lessons at this site, and, if we were to actually try this time, then it would be difficult to know how far back to begin with you.)
The limit that you posted is the derivative of cos(x) at x = a.
Are you familiar with the following limit definition for the derivative?
\(\displaystyle \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\)
This is how I start.
Then I use the following limits. Are you familiar with them?
\(\displaystyle \lim_{h \to 0}\frac{sin(h)}{h}\)
\(\displaystyle \lim_{h \to 0}\frac{cos(h) - 1}{h}\)
Cheers,
~ Mark
Hello DJO:
Please post some clues as to what you already know about these types of exercises. Show any work and reasoning, if you can.
If you already understand the question, then are you able to say something about why you're stuck?
(Please read the post titled, "Read Before Posting", if you have not already done so. We do not, in general, provide lessons at this site, and, if we were to actually try this time, then it would be difficult to know how far back to begin with you.)
The limit that you posted is the derivative of cos(x) at x = a.
Are you familiar with the following limit definition for the derivative?
\(\displaystyle \lim_{h \to 0}\frac{f(x + h) - f(x)}{h}\)
This is how I start.
Then I use the following limits. Are you familiar with them?
\(\displaystyle \lim_{h \to 0}\frac{sin(h)}{h}\)
\(\displaystyle \lim_{h \to 0}\frac{cos(h) - 1}{h}\)
Cheers,
~ Mark
Re: Calculus Limits
This question is really a 'somnolence' test. I.E. it tests whether you are asleep. If you are awake, then you would notice that:
1. The definition of derivatve at x = a is:
f'(a) = lim[as x->a] (f(x) - f(a))/(x-a)
2. Your expression:
Lim[as x->a] (cos(x)-cos(a))/(x-a)
bears a striking resemblance to that definition.
Are you awake yet?
djo201 said:
This question is really a 'somnolence' test. I.E. it tests whether you are asleep. If you are awake, then you would notice that:
1. The definition of derivatve at x = a is:
f'(a) = lim[as x->a] (f(x) - f(a))/(x-a)
2. Your expression:
Lim[as x->a] (cos(x)-cos(a))/(x-a)
bears a striking resemblance to that definition.
Are you awake yet?