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I am having trouble with this problem:

Use the definition of limit to prove that: $$\lim_{x\to \infty} \frac{\sin x}{x(\sin x)^2 +1} = \,?$$

I have concluded that the limit must be 0, but I am having trouble proving it.

Using the limit definition, I must show that

$$\left\lvert \frac{\sin x}{x(\sin x)^2 +1}- 0\right\rvert < \epsilon$$

Cheyenne
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  • Just for future reference, \infty shows $\infty$ and \sin x shows $\sin x$. – Em. Jan 15 '16 at 06:16
  • Why do you think it must be 0? (If you are precise you can turn your answer into a proof) – mathematician Jan 15 '16 at 06:24
  • @mathematician If $x=n\pi$ the expression is $0$. So it is clear that if the limit exists it must be $0$. But proving the limit exists might be tricky... :-) – David Jan 15 '16 at 06:26
  • I am assuming it must be zero because as x increases, the x in the denominator dominates and the function sort of "wiggles" around 0 until it eventually becomes zero. I am having trouble putting it into a mathematical statement. – Cheyenne Jan 15 '16 at 06:31

2 Answers2

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Clearly we may assume that $x>0$. Consider two cases:

  • if $|\sqrt x\sin x|<1$ then $|\sqrt x\sin x|<x\sin^2x+1$;
  • if $|\sqrt x\sin x|\ge1$ then $|\sqrt x\sin x|\le x\sin^2x<x\sin^2x+1$.

Hence in all cases we have $$\Bigl|\frac{\sin x}{x\sin^2x+1}\Bigr|<\frac1{\sqrt x}\ ;$$ so given $\varepsilon>0$, taking $x>1/\varepsilon^2$ guarantees that the LHS is less than $\varepsilon$. I'll leave you to turn this into a formal proof.

David
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This is constructed a special way. That is key to the solution. It is an example of a function that can be shown to converge to zero as x approaches infinity by two separate sequences on the function +1/x and -1/x. This values are taken by the function under consideration is sine is 1 and therefore the square of sine if 1.

These are neither maxima nor minima of the function but relative ones. The combination of x and sine makes it hard to calculate the values. The function is close to the periodicity of sine but drops like 1/x, -1/x to zero.

function with first derivative and inner bound guide curve

and on a longer intervall

function with first derivative and inner bound guide curve

This does not imply that +/-1/x is minorant or majorant to the function. Both are steady and infinitely differentiable and therefore close related in the graphs.

The function is limit on the positive reals since the value at 0 is 0. There is a maximum at 0.816834.

There are crossing points to -/+1/x by the functions on the positive reals. These start with x bigger than 2 .

minimum crossing

They take place arount the maxima and minima of the sine. With that bounds to the crossings it is assured that the function decreases to 0 for x-> ∞. All these are numerical accessible value like the first maximum they are close to a transcendental number /4.

The series of crossings starts with

{{x -> 6.47523}, {x -> 7.27203}, {x -> 8.38635}, {x -> 9.30196}, {x ->
    12.653}, {x -> 13.7369}, {x -> 14.5253}, {x -> 15.6393}, {x -> 
   18.9056}, {x -> 20.0948}, {x -> 20.7404}, {x -> 21.9433}, {x -> 
   25.1742}, {x -> 26.4219}, {x -> 26.9821}, {x -> 28.2376}}

and for each natural positive n with n + there is another crossing. If it would not be there the equation 1/2 (2 - x + x Cos[2 x] + 2 x Sin[x])=0 would have no solution. That can not be because the trigonometric functions are periodic.

Steffen Jaeschke
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