# Is WolframAlpha wrong? or am I?

Problem :

Determine a limit $$\lim_{x\to\infty}xe^{\sin x}$$

exists or not. If it exists, find a limit.

Since $$-1\le \sin x \le 1$$, I can say that $$\displaystyle0<\frac{1}{e}\le e^{\sin x}\le e$$.

Multiply $$x$$ both side, $$\displaystyle0<\frac{x}{e}\le xe^{\sin x}\le ex$$

Since $$\displaystyle\lim_{x\to\infty} \frac{x}{e}=\infty$$, I think $$\displaystyle\lim_{x\to\infty}xe^{\sin x}=\infty$$ also.

But, WA says it is indeterminate form :

Am I correct? or Am I wrong?

If I'm wrong, where did I make a mistake?

• I agree that the limit is $\infty$, and I agree with your reasoning.
– lulu
Aug 3 at 11:45
• WrongAnswer should not be trusted unless you also have a proof. It is trivial for any mathematically trained person to come up with a limit question that WA answers completely wrongly. Don't be surprised if within the next few weeks WA is modified so that it says the correct answer when tested on your question here. That doesn't mean that they fixed the bug; I've seen what they did, which is just to hard-code some cases and call it a day. That's why my second sentence will still hold in the future. Aug 3 at 20:05
• This may be a bug in Mathematica's Limit function. You should consider sending in feedback so they can address it. Aug 3 at 20:22
• @user170231: They usually don't address bugs in mathematically correct ways. By reporting these, it is likely that it would become even worse, because the code will grow longer just in the sense of hard-coded cases. Aug 3 at 21:23
• I can confirm that this is a bug in Mathematica 13.0: the input Limit[x Exp[Sin[x]], x -> Infinity] wrongly produces Indeterminate, even if you specify Direction -> "FromBelow". Aug 4 at 7:56

No, you are perfectly right.

Maybe what WolframAlpha is telling you is that it could not resolve the limit. Just because you are smarter than the computer does not mean you did something wrong.

Although the WA algorithm is almost advanced, you caught a gap there. However, WA calculates the limit $$\lim_{x\to\infty} \frac{1}{x e^ {\sin x}}=0$$ correctly. This justifies the original limit.

EDİT:

You multiplied both side of the equation $$0<\frac{1}{e}\le e^{\sin x}\le e$$ by $$x$$. So, you wrote that $$0<\frac{x}{e}\le xe^{\sin x}\le ex$$ It would be a more rigorous solution if you just stated that you did this with the $$x>0$$ condition.

• Minor nitpick: knowing that 1/f(x) tends to zero doesn't in itself prove that f(x) tends to +infinity. Consider e.g. f(x) = x*(-1)^floor(x). But in this particular case, the gap is easily closed by observing that the function is non-negative. Aug 4 at 2:05

Using Maple, the limit is calculated correctly ($$\displaystyle\lim_{x\to\infty}xe^{\sin x}=\infty$$)

But when I use "limit methods" to ask for clarification, it tries to do this:

$$\displaystyle\lim_{x\to\infty}xe^{\sin x}$$

$$=\displaystyle(\lim_{x\to\infty}x)(\lim_{x\to\infty}e^{\sin x}) [product]$$

$$=\displaystyle\infty(\lim_{x\to\infty}e^{\sin x}) [identity]$$

$$=\displaystyle\infty(e^{\lim_{x\to\infty}{\sin x}}) [exp]$$

$$=\displaystyle{undefined} [sin]$$

$$\displaystyle\lim_{x\to\infty}xe^{\sin x}={undefined}$$

Maybe, Wolfram Alpha tries to do something similar.

And the explanation provided for $$\lim_{x\to\infty} \frac{1}{x e^ {\sin x}}=0$$ is:

$$\lim_{x\to\infty} \frac{1}{x e^ {\sin x}}$$

$$\displaystyle=1/{\lim_{x\to\infty} \frac{1}{x e^ {\sin x}}} [power]$$

$$\displaystyle=1/{(\lim_{x\to\infty}x)(\lim_{x\to\infty}e^{\sin x})} [product]$$

$$\displaystyle=0 [identity]$$

$$\lim_{x\to\infty} \frac{1}{x e^ {\sin x}} = 0$$