Limit question- Urgently need a solution

[Abraham]

Calculus - Limits question- Urgently need a solution

I have corrected the questions. this is where I got stuck.
I would be very glad if someone can help me with this Limit problem.

1.lim{x->0) sin(x)(1-cos(x))/2x^8 =
lim{x->0)sin(x)/2x^8*1-cos(x) = 1/2lim(x->0)sin(x)/x^8*1-cos(x)


2.
lim{x->0) xsin(x)/1-cos(x) =lim(x->0)x/1-cos(x)*sin(x)


Thank you very much for your help

 

[Deleted member 4993]

I would be very glad if someone can help me with this Limit problem.

1. sin(x)(1-cos(x))/2x^8

2. xsin(x)/1-cos(x)

Thank you very much for your help

What are your thoughts?

Please share your work with us ...even if you know it is wrong

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[Jedi Equester]

You missed to tell us, where you are approaching to. Also I assume for at least the last one you missed some brackets.

2. xsin(x)/(1-cos(x))

If we assume we are interested in x->0 you might think of l'Hospital.

 

[stapel]

I would be very glad if someone can help me with this Limit problem.

1. sin(x)(1-cos(x))/2x^8

2. xsin(x)/1-cos(x)

You say "this" limit problem, which suggests that the two expressions above relate to one exercise. But what was that exercise? What were its instructions? In particular, what is the "limit" for this "problem"? (Neither of the two expressions is a limit, having no "limit" statement or target value, such as "x getting very close to 0 from the left".)

When you reply, please include all information necessary to this one exercise, along with a clear statement of your thoughts and efforts so far, so we can see where you're having trouble. Thank you!

 

[MGM]

For no. 2, multiply both numerator and denominator by (1+cos x).
Then, a (1-cos²x)) term comes, which you can convert to sin²x.

So, you are left with x(1+cos x)/sin x, which makes things much simpler.

 

[stapel]

...what was that exercise? What were its instructions? In particular, what is the "limit" for this "problem"? (Neither of the two expressions is a limit, having no "limit" statement....)

When you reply, please include all information...along with a clear statement of your thoughts and efforts so far....

I have corrected the questions.

In future, kindly please reply with corrections, clarifications, etc, so that the (literal) thread of the conversation is not lost.

this is where I got stuck.
I would be very glad if someone can help me with this Limit problem.

1.lim{x->0) sin(x)(1-cos(x))/2x^8 =
lim{x->0)sin(x)/2x^8*1-cos(x) = 1/2lim(x->0)sin(x)/x^8*1-cos(x)

I will guess that the original exercise was the following:

. . . . .\(\displaystyle \mbox{1. Find the following: }\, \)\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\, \)\(\displaystyle \dfrac{\sin(x)\, (1\, -\, \cos(x))}{2x^8}\)

Then your "work done" was this:

. . . . .\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\,\)\(\displaystyle \dfrac{\sin(x)}{2x^8\, (1\, -\, \cos(x))}\, =\, \dfrac{1}{2}\,\)\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, 0}\,\)\(\displaystyle \dfrac{\sin(x)}{x^8\, (1\, -\, \cos(x))}\)

But this makes no sense, so maybe some grouping symbols went missing somewhere...? Also, what did you do after taking the "2" from the denominator out front? What trig limits, etc, did you bring to bear?

2.
lim{x->0) xsin(x)/1-cos(x) =lim(x->0)x/1-cos(x)*sin(x)

I'm sorry, but I don't follow. How is the sine going from the numerator (in the first "limit" statement) to the denominator (in the second "limit" statement)? What steps came in between? What was your reasoning?

Please be complete. Thank you!