Evaluating A Limit with Tangent and Sine

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SyberMath

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Idk why this trick works, but using L’Hopitals rule allows us to define the limit at x=3

Edit: ok so this works due to the Taylor approximation of sin and tan, tho this wouldn’t work all the time unless you adjust the Taylor series to start from that point

TheFireBrozTFB

I would say this is also solvable by reaching the point where we can apply the lim[x->0] senx/x

babyboy

Direct-sub gives form 0/0, which authorizes us to use L'Hopital's Rule: lim f(x)/g(x) = lim f '(x)/g'(x).
In this case, taking derivative of top and bottom separately, using chain rule as needed, you get lim [cos(x^2 - 3x)]*[2x-3] / {2[sec(2x-6)]^2} = (2x-3)/2 = 3/2.
[since lim cos(A) = 1 & lim sec(A) = 1 as A->0].
I have such an unreliable memory, I'd be real worried about trying to correctly remember yet another trick of limited application. (pun intended! ;-)

timeonly

Please let me know if this is correct or not
lim sin(x²-3x)/tan(2x-6)
x–>3
=lim sin(x(x-3))/tan(2(x-3))
x–>3
= lim
x–>3
=lim cos(2(x-3)) . lim sin(x(x-3))/lim sin(2(x-3))
x–>3 x–>3. x–>3
Since there is no issue with cos (2(x-3)), so we can put x=3
So cos(2(3-3))=cos0=1
Now,
lim sin(x(x-3))/lim sin(2(x-3))
x–>3. x–>3
Now considering x-3=a =>x=a+3
When x–>3, a–>0
So, we can conclude
lim sin((a+3)a)/lim sin2a
a–>0 a–>0
lim sin(a(a+3))(a+3)/a(a+3)/2 lim
a–>0 sin2a/2a
lim (a+3) . lim sin(a(a+3))/a(a+3)/
a–>0 a–>0 2 lim sin2a/2a
a–>0
= (0+3) ×1×1/(2×1)
(Since we know lim sinx/x =1)
x–>0
=3/2
Answer:3/2
Hope it will be correct

astrodude

Shouldn't x^2-3x= 2x-6=0. So the equation would become sin(0) /tan(0) which further simplifies to 0/0 which is undefined??

imbisatazim

This problem is also possible with L'Hopital's theorem. ^^

hippo

Используя следствия из I замечательного предела, можно устно получить ответ 3/2.

КатяРыбакова-шд

L'hospital rule is a better approach...

parthtomar

Does this work for different powers of sin and tan?

aymanabdellatief

3/2 il sinx e la tgx sono approssimabili a x per x>0... Perciò x(x-3)/2(x-3)=x/2...

giuseppemalaguti

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