I am having trouble finding the values of a and b for which limx→0\lim \limits_{x \to 0} (\frac{\sin (ax) + bx)} {(x^3)}) = -32/3 (\frac{\sin (ax) + bx)} {(x^3)}) = -32/3 . In Mathematica I have entered:

f[x_] := (Sin[a * x] + (b * x))/x^3

Solve[Limit[f”'[x], x -> 0] == -32/3, {a, b}]

The hint for this problem is that the third derivative of the function should be taken. Mathematica gives back to me a set of empty curly braces. Could someone help me understand why what I am doing is incorrect?

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f[x_] := (Sin[a*x] – (a*x))/x^3; Limit[f[x], x -> 0]

– Dr. belisarius

Oct 15 ’15 at 18:40

From f[x_] := (Sin[a*x] + (b*x))/x^3, Series[f@x, {x, 0, 3}] you may see that b == -a

– Dr. belisarius

Oct 15 ’15 at 18:42

3

Kinda’ cheap, but: In[90]:= SolveAlways[ Normal[Series[(Sin[a*x] + (b*x))/x^3, {x, 0, 1}]] == -32/3, x] Out[90]= {{a -> 4, b -> -4}, {a -> 2 (-1 – I Sqrt[3]), b -> 2 (1 + I Sqrt[3])}, {a -> 2 (-1 + I Sqrt[3]), b -> 2 (1 – I Sqrt[3])}}

– Daniel Lichtblau

Oct 15 ’15 at 18:46

…and @Daniel’s approach is effectively equivalent to l’Hôpital.

– J. M.♦

Oct 15 ’15 at 19:06

Amplifying on comment by @DanielLichtblau, Cases[SolveAlways[Normal[Series[(Sin[a*x] + (b*x))/x^3, {x, 0, 1}]] == -32/3, x], _?(FreeQ[#, Complex] &)] returns {{a -> 4, b -> -4}}

– Bob Hanlon

Oct 15 ’15 at 20:46

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1 Answer

1

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Let

f[x_, a_, b_] := (Sin[a*x] + (b*x))/x^3

and look for the limit in question

Limit[f[x, a, b], x -> 0]

(*

Out[10]= DirectedInfinity[a + b]

*)

It has a finite value only if a + b == 0.

Hence we consider

Limit[f[x, a, -a], x -> 0]

(*

Out[12]= -(a^3/6)

*)

Now this has to go into the equation to be solved.

With the result

Solve[% == -32/3]

(*

Out[13]= {{a -> 4}, {a -> -4 (-1)^(1/3)}, {a -> 4 (-1)^(2/3)}}

*)

Hence the real solution is {a,b} = {4,-4}, the others correspondingly.

Have I missed something or have you? It seems to me the questions asks for the limit of the 3rd derivative (f”’), not of f itself.

– m_goldberg

Oct 15 ’15 at 22:32

@m_goldberg – the OP asks for the limit of the function; however, the OP also states that her/she was given a hint to take the third derivative. This hint apparently led the OP to attempt to solve an equation involving the limit of the third derivative of the function. The answers and comments have chosen to show approaches to get the requested limit of the function without making any use of the hint, i.e., without making any use of f”'[x].

– Bob Hanlon

Oct 15 ’15 at 23:43

@BobHanlon. Ah, so I’m the one who missed the point. Thanks for straightening me out.

– m_goldberg

Oct 16 ’15 at 3:21