Limit using taylor

[Anakin_99]

Hi guys I was wondering if is this limit correct, there is no solution on my book and I m not sure so... here it is

 

[BigBeachBanana]

I can't follow your work. Seem like you're skipping a lot of steps. Where's the Taylor series?

 

[Anakin_99]

I can't follow your work. Seem like you're skipping a lot of steps. Where's the Taylor series?

taylor series is for the exponencial aka e^x = 1 + x + (x^2)/2! + (x^3)/3! + .....

 

[BigBeachBanana]

taylor series is for the exponencial aka e^x = 1 + x + (x^2)/2! + (x^3)/3! + .....

1) Nowhere you show that in your solution
2) Where did the [imath](1+h)[/imath] go from the first line?
3)Where is the Taylor's expansion for [imath]\sin h(h+2)?[/imath]

Your answer is incorrect. Please show your work in more detail.

 

[Anakin_99]

1) Nowhere you show that in your solution
2) Where did the [imath](1+h)[/imath] go from the first line?
3)Where is the Taylor's expansion for [imath]\sin h(h+2)?[/imath]

Your answer is incorrect. Please show your work in more detail.

1) ok I forgot to tell u which taylor exp I used, i thought it was obvius
2) (1+h) is negligible cause h--->0 so basically it s a multiplication for 1
3) sin [h(h+2)] = sin (2h) cause h--->0 and 2+0 = 2, then sin (2h) is asymptotic to 2h for h---->0

Anyway if u think my answer is incorrect please explain me cause I didn t get it

 

[BigBeachBanana]

First, rewrite as follow:
[math]\lim_{h \to 0^-} \frac{\frac{e^h-1}{(1-h)(\sin [h(h+2)]}-0.5}{h} =\lim_{h \to 0^-} \frac{1}{h}\left({\frac{e^h-1}{(1-h)(\sin [h(h+2)]}-0.5}\right)[/math]You know the expansion for [imath]e^{h}[/imath], but what's the Taylor's series for [imath]\sin (h^2+2h)?[/imath]

 

[Anakin_99]

First, rewrite as follow:
[math]\lim_{h \to 0^-} \frac{\frac{e^h-\frac{1}{2}}{(1-h)(\sin [h(h+2)]}}{h} =\lim_{h \to 0^-}\frac{1}{h}\left(\frac{e^h-\frac{1}{2}}{(1-h)\sin (h^2+2h)}\right)[/math]You know the expansion for [imath]e^{h}[/imath], but what's the Taylor's series for [imath]\sin (h^2+2h)?[/imatI [/QUOTE][/imath]

First, rewrite as follow:
[math]\lim_{h \to 0^-} \frac{\frac{e^h-\frac{1}{2}}{(1-h)(\sin [h(h+2)]}}{h} =\lim_{h \to 0^-}\frac{1}{h}\left(\frac{e^h-\frac{1}{2}}{(1-h)\sin (h^2+2h)}\right)[/math]You know the expansion for [imath]e^{h}[/imath], but what's the Taylor's series for [imath]\sin (h^2+2h)?[/imath]

I used asymptotic:
sin (h^2 + 2h) = sin [h(h+2)] = sin (2h) that is asymptotic to 2h cause h--->0 aka 2h--->0

 

[BigBeachBanana]

I used asymptotic:
sin (h^2 + 2h) = sin [h(h+2)] = sin (2h) that is asymptotic to 2h cause h--->0 aka 2h--->0

But the point is to use Taylor's? The answer is -1/2.

 

[Anakin_99]

But the point is to use Taylor's? The answer is -1/2

You can use whatever u want not strictly taylor, why u say answer is -1/2 ?
Anyway asymptotics are taylor approximations at first order

 

[BigBeachBanana]

You can use whatever u want not strictly taylor, why u say answer is -1/2 ?
Anyway asymptotics are taylor approximations at first order

Your title of the thread "limit using Taylor"? Maybe make it clearer with instruction.
If you were to to expand using Taylor series, you can see that:
[math]f(h)=-\frac{1}{2}+ \frac{11h}{12} -\frac{29h^2}{48}+\dots[/math]

 

[Anakin_99]

Your title of the thread "limit using Taylor"? Maybe make it clearer with instruction.
If you were to to expand using Taylor series, you can see that:
[math]f(h)=-\frac{1}{2}+ \frac{11h}{12} -\frac{29h^2}{48}+\dots[/math]

1) I did use taylor in both cases, I repeat, asymptotics are taylor approximations at first order so yes, i am using taylor in any case
2) what are you expanding in your answer I didn t get where those numbers come out from
Please just tell me where Im wrong or post ur steps if u prefer

 

[BigBeachBanana]

1) I did use taylor in both cases, I repeat, asymptotics are taylor approximations at first order so yes, i am using taylor in any case
2) what are you expanding in your answer I didn t get where those numbers come out from
Please just tell me where Im wrong or post ur steps if u prefer

Where you're wrong is assuming (1+h) is negligible and asymptomatic of [imath]\sin(h^2+2h)=2h[/imath]
Graph each of the functions, starting from the original to the end. You can see that you're not evaluating the limit of the same function, thereby the limits at 0 also change after every step of your simplifications.

 

[Anakin_99]

Where you're wrong is assuming (1+h) is negligible and asymptomatic of [imath]\sin(h^2+2h)=2h[/imath]
Graph each of the functions, starting from the original to the end. You can see that you're not evaluating the limit of the same function, thereby the limits at 0 also change after every step of your simplifications.

That part was done with my calculus prof at university, now I explain:
1) h--->0 so 1+0 = 1 right ? this means that (1+h) * sin (...) = 1 * sin (...) ok ? And thats why (1+h) disappears.
2) sin (h^2 + 2h) = sin [ h ( h +2 ) ] so if h--->0 it can be removed from the internal ( ) cause 2 + 0 = 2 right ? So it becomes sin ( 2h ) that is asymptotic to 2h and that' s all.

 

[BigBeachBanana]

That part was done with my calculus prof at university, now I explain:
1) h--->0 so 1+0 = 1 right ? this means that (1+h) * sin (...) = 1 * sin (...) ok ? And thats why (1+h) disappears.
2) sin (h^2 + 2h) = sin [ h ( h +2 ) ] so if h--->0 it can be removed from the internal ( ) cause 2 + 0 = 2 right ? So it becomes sin ( 2h ) that is asymptotic to 2h and that' s all.

You clearly didn't follow my suggestion and graph the functions, so I have no incentive to help you. If that's what your professor told you to do that for limits then he/she shouldn't be teaching. This is all I can show you. Observe the value at h=0.

Your original function and its plot:


Your simplified function:

 

[Anakin_99]

That's not a function thats a limit its different, plot it with a graph doesnt help.
It s like to say that lim x--->0 sin(x) is not asymptotic to x cause the graph is different, u understand this ?

 

[BigBeachBanana]

That's not a function thats a limit its different, plot it with a graph doesnt help.
It s like to say that lim x--->0 sin(x) is not asymptotic to x cause the graph is different, u understand this ?

Limit is the value that a function approaches as the input approaches some value. i.e. [imath]h \to 0[/imath] of [imath]f(h)[/imath]. In other words, you're trying to find out around h=0, what is the function approaching. By graphing the 2 functions, you can clearly see that [imath]f(h)[/imath] around h=0 are approaching 2 complete different values. Ergo, your simplification is wrong. I think you need to review what limit means.

 

[Anakin_99]

Limit is the value that a function approaches as the input approaches some value. i.e. [imath]f \to 0[/imath] of [imath]f(h)[/imath]. In other words, you're trying to find out around h=0, what is the function approaching. By graphing the 2 functions, you can clearly see that [imath]f(h)[/imath] around h=0 are approaching 2 complete different values. Ergo, your simplification is wrong. I think you need to review what limit means.

So you clearly dont know what an asymptotic is, I m not gonna lose other time with u, luckly my prof answered my mail and told me the limit is correct, anyway my university is the 1° in my country and 9° in europe and prof is really smart. If u dont understand why (h+1) * .... is irrelevant for h--->0 or why sin(h(h+2)) ~ 2h, I cant help u dude, have a nice day.

 

[BigBeachBanana]

So you clearly dont know what an asymptotic is, I m not gonna lose other time with u, luckly my prof answered my mail and told me the limit is correct, anyway my university is the 1° in my country and 9° in europe and prof is really smart. If u dont understand why (h+1) * .... is irrelevant for h--->0 or why sin(h(h+2)) ~ 2h, I cant help u dude, have a nice day.

Tell your professor that WolframAlpha disagrees. He will reconsider his answer.