Question

Any help with an explanation to the following problem would be appreciated:

Consider the limit as x approaches 0 for (sin 2x - 2x)/x^3. Graphically, the solution is -4/3. Using L'Hopital's rule also results in the solution of -4/3. However, analytically, lim(x->0) [sin 2x/x^3 - 2x/x^3] = lim (x->0) [(2x*sin 2x)/(2x * x^3) - 2/x^2] = lim(x->0) [2/x^2 - 2/x^2] = lim(x->0) 0 = 0.

Why is the analytic solution different from the graphical and L'Hospital methods?

Answer 1

You can't take sin 2x /x to the limit without taking the rest also, you can only evaluate everything at once.

Answer 2

@ivanmlerner
Is it not true that the limit of a difference is the difference of the limits?

Answer 3

Yes, that is true, but I'm talking about the multiplication and division. When you multiply by 2x/2x the next step you did was taking the limit of (sin y)/y = 1 to get rid of the sin right? But you can't do that without evaluating the 1/x^2 that multiplies it.

Answer 4

limit of a difference is the difference of the limits but not when u have 0-0

Answer 5

thats something like 0/0 or 0^0

Answer 6

So, you're saying lim (x->0) 2/x^2 is infinity, so this limit is nonexistent.
The limit that I'm taking is an indeterminate form: "infinity minus infinity."

Okay. So, how can I do this limit analytically (correctly) and not use L'Hospital's rule?

Answer 7

try expanding sin(2x) using Taylor series

Answer 8

No if that were the limit: 2/x^2 - 2/x^2 you can do it, the problem is when you do the limit of the sin x /x, but I tried doing this limit and didn't get anywere, I only know what is wrong, not how to do it so far.

Answer 9

sorry ur case is infty-infty

Answer 10

and yeah as exper said taylor series

Answer 11

\[ {\sin (2x) - 2x \over x^3} \times {\sin (2x) + 2x \over \sin (2x) + 2x
} ={\sin^2 (2x) - 4x^2 \over 2x^4 } \times {2x \over \sin (2x) + 2x} =
\\
{-4\sin^4x + 4 \sin^2x - 4 x^2 \over 4 x^4} = -1+ {\sin^2x - x^2 \over x^4 }=-1 + 2 \times {L \over 8} \\
L - {L \over 4} = -1 \implies L = {-4 \over 3} \]

Answer 12

lol i made a bad mistake somewhere

Answer 13

Does anyone think the type reasoning given below is mathematically sound?
===========================

Think of this problem and other limit at zero problems by considering the quadratic approximations.

sin(2x) behaves like 2x - (2x)^3/6 near x = 0, so (sin 2x - 2x)/x^3 behaves like (2x - 4x^3/3 -2x)/x^3 = 4/3 near x = 0.

Answer 14

probably ... try making use of WA ... i use it violently
http://www.wolframalpha.com/input/?i=limit+x-%3E+0+%28sin^2x+-+x^2%29%2Fx^4

I think that would be fine.

Answer 15

Thanks, everyone, for your work on this problem.