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Evaluate the following limit

$$\lim_{n \rightarrow \infty}{\left(n^2 - \frac{1}{\sin^2(\frac{1}{n})} \right)}$$

From wolfram alpha, answer is $-\frac{1}{3}$. From here, we obtain the answer by using either Taylor series or L'Hopital rule. I try to apply L'Hopital rule to this question. I obtain

$$\lim_{n \rightarrow \infty}{\left(n^2 - \frac{1}{\sin^2(\frac{1}{n})} \right)}=\lim_{n \rightarrow \infty}{\frac{n^2\sin^2(\frac{1}{n})-1}{\sin^2(\frac{1}{n})}} =\frac{-1}{0}$$

What's wrong with my working?

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  • $\begingroup$ Notice that $n^2\sin^2(1/n)$ does not necessarily approach to $0$ as $n$ tends to infinity, because $n^2\rightarrow\infty$ as $n$ tends to infinity while $\sin^2(1/n)\rightarrow 0$. $\endgroup$ – Katie Imach Oct 25 '16 at 1:29
  • $\begingroup$ @KatieImach: Good catch. I applied wrong fact. $\endgroup$ – Idonknow Oct 25 '16 at 1:31
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    $\begingroup$ Please see my answer to this question math.stackexchange.com/questions/1976711/… It is the same problem letting $x=\frac{1}{x}$. $\endgroup$ – Rene Schipperus Oct 25 '16 at 2:35
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Use substitution $y:=\frac{1}{x}.$ Then $x\to+\infty \implies y\to0.$ So, $$\lim_{n\to+\infty}\Biggl(n^2-\frac{1}{\sin^2\bigl(\frac{1}{n}\bigr)}\Biggr)=\lim_{y\to0}\Biggl(\frac{1}{y^2}-\frac{1}{\sin^2(y)}\Biggr) = \lim_{y\to0}\Biggl(\frac{\sin^2(y)-y^2}{y^2\sin^2(y)}\Biggr).$$ Applying L'Hopital rule 4 times yields the following limit: $$\lim_{y\to0}\Biggl(\frac{\sin^2(y)-y^2}{y^2\sin^2(y)}\Biggr) = \lim_{y\to0}\Biggl(\frac{-4\cos(2y)}{12\cos(2y)-16y\sin(2y)-4y^2\cos(2y)}\Biggr).$$ Since $$\begin{align} -4\cos(2y)\rightarrow -4, && (y\to0) \\ 12\cos(2y)\rightarrow12, && (y\to0) \\ -16y\sin(2y)\rightarrow0, && (y\to0) \\ -4y^2\cos(2y)\rightarrow 0, && (y\to0) \end{align}$$ the last limit is equal to $$\lim_{y\to0}\Biggl(\frac{-4\cos(2y)}{12\cos(2y)-16y\sin(2y)-4y^2\cos(2y)}\Biggr) = -\frac{4}{12} = -\frac{1}{3}.$$

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