$$\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=\lim_{x\to 0 }\frac{x^2\sin\frac{1}{x}}{x}=\lim_{x\to 0} x\sin\frac{1}{x}=0$$
Is this solution right?
Thank you very much!
Asked
Active
Viewed 94 times
4
-
1It would help to know how you progressed from one step to the next. Particularly from the first to the second equality. – PrincessEev Oct 16 '19 at 01:33
-
Firstly, you mean $\lim_{x \to \infty}$, not $\lim_{n \to \infty}$ right? Secondly, where did the $\sin$ go from first to the second equality? – Clement Yung Oct 16 '19 at 01:33
-
sorry, it is $x\to 0.$ – Jacob.Lee Oct 16 '19 at 01:35
-
@ClementYung Probably $\sin(x)\leq x$ (for $x\geq 0$) was used. – Cornman Oct 16 '19 at 01:38
-
@Cornman good call, thanks. Jacob, assuming you meant inequality in the first equality, the approach is about right. Instead of evaluating directly, consider squeeze theorem. – Clement Yung Oct 16 '19 at 01:41
3 Answers
5
I assume you're using that $\sin (r)\approx r$ in your first step. While this is a good heuristic, it does not constitute a rigorous argument.
Perhaps try something like
$$ \left|\frac{\sin(x^2\sin\frac{1}{x})}{x}\right|\le \frac{x^2\left|\sin\frac{1}{x}\right|}{|x|}\le |x| $$ and the RHS tends to $0$ as $x\to 0$.
Reveillark
- 13,044
-
@Reveilark I used the Infinitesimal equivalent substitution technique: $\sin x\sim x$ as $x\to 0.$ Is it wrong? – Jacob.Lee Oct 16 '19 at 01:54
-
-
-
@Jacob I think what you did is fine. Really you've just multiplied top and bottom by the argument of the sine, split the limit up into two finite products, on of which is well understood to be one. You just have to be maybe slightly more careful about explaining your justification of that step. – Cade Reinberger Oct 16 '19 at 02:46
0
$$\lim_{x\to 0}|\frac{\sin(x^2\sin\frac{1}{x})}{x}|=$$
$$\lim_{x\to 0 }|\frac{\sin(x^2\sin\frac{1}{x})}{x^2} (x)|\le $$
$$\lim_{x\to 0 }|\frac{(x^2\sin\frac{1}{x})}{x^2} (x)| =$$
$$\lim_{x\to 0}| x\sin(\frac{1}{x})|=0.$$
Mohammad Riazi-Kermani
- 68,728
-
Let $x_n=\frac{1}{n\pi}$. $x_n=\frac{1}{n\pi}\to 0~~ (n\to \infty)$. The denominator $x^2sin(\frac{1}{x})$ will be zero at $x_n$. It makes nonsense. – Jacob.Lee Oct 16 '19 at 02:11
-
@Dr.Jacob.Z.Lee: your own approach suffers from the same defect. The replacement of $\sin x$ with $x$ while evaluating a limit as $x\to 0$ involves the use of divide and multiply by $x$ followed by replacing the ratio $(\sin x) /x$ with its limit $1$. Ideally this step should not be implicit but rather explicitly shown (or at least mentioned in words). – Paramanand Singh Oct 18 '19 at 02:55
0
A simple proof: $$ \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x} = \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x} \cdot\frac{x}{x}\frac{\sin(\frac{1}{x})}{\sin(\frac{1}{x})} = \lim_{x\rightarrow0} \frac{\sin(x^2\sin(\frac{1}{x}))}{x^2\sin(\frac{1}{x})} \cdot x\sin(\frac{1}{x}) $$ The term $\frac{\sin(x^2\sin(\frac{1}{x}))}{x^2\sin(\frac{1}{x})} $ goes to $1$ as $ x \rightarrow 0 $ by the famous identity $\lim_{f(x)\rightarrow0}\frac{\sin(f(x))}{f(x)} = 1 $ as the term $ \sin(1/x) $ is bounded between [-1,1] and the term $ x^2 $ goes to zero, so overall the dummy function $ f(x) = x^2\sin(\frac{1}{x}) \rightarrow 0 $ as $ x \rightarrow 0 $. We now have: $$ \lim_{x\rightarrow0} x\sin(\frac{1}{x}) = 0$$ as again the term $ \sin(1/x) $ is bounded between [-1,1]. I hope the answer is satisfying.
Dinno Koluh
- 753
-
This is an expanded version of what is presented in the question. I am not sure if the asker is aware if his methods actually means this expanded version. But both of these have the a subtle flaw. You can't bring in $\sin(1/x)$ in denominator as it vanishes many times as $x\to 0$. The right approach is to deal using inequalities. – Paramanand Singh Oct 18 '19 at 02:49
-
@Singh I agree with you that there is a flaw in his solution. As I said above, when $x=\frac{1}{n\pi}$, $\sin\frac{1}{x}$ will vanish. – Jacob.Lee Oct 18 '19 at 23:40