# $\lim_{n\to\infty}\frac{e^{\frac{1}{n^2}}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}$

$$\lim_{n\to\infty}\frac{e^{1/n^2}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}, \quad n \in \mathbb{N}$$

I'm stuck trying to evaluate this limit. I have tried applying l'Hôpital's rule to the corresponding function in $$\mathbb{R}$$, but even after differentiating all that stuff I still end up with a $$\frac{0}{0}$$ form. There has to be something I'm missing. Any hints?

Hint. Try by using Taylor expansions. As regards the numerator $$e^{\frac{1}{n^2}}-\cos \frac{1}{n}=1+\frac{1}{n^2}+o(1/n^2)- \left(1-\frac{1}{2n^2}+o(1/n^2)\right)=\frac{3}{2n^2}+o(1/n^2).$$ Whereas, for the denominator we have that $$\frac{1}{n}\log(1+\frac{1}{n})-(\sin\frac{2}{n})^2=?$$ Can you take it from here?

• The denominator becomes $$\frac{1}{n}(\frac{1}{n}+o(\frac{1}{n}))-\frac{4}{n^2}+o(\frac{4}{n^2}) = \frac{1}{n^2}+o(\frac{1}{n})-\frac{4}{n^2} + o(\frac{4}{n^2})= -\frac{3}{n^2}+o(\frac{1}{n^2})$$ hence I end up with $$\frac{\frac{3}{2n^2}+o(\frac{1}{n^2})}{-\frac{3}{n^2}+o(\frac{1}{n^2})}\to-\frac{1}{2}$$ Correct? – Samuele B. Jan 2 at 15:32
• @SamueleB. Yes, you are correct! – Robert Z Jan 2 at 15:36

HINT

Write out the first couple of terms of Taylor series for each of the 4 pieces, it will become clear.

For example, $$e^{1/n^2} = 1 + \frac{1}{n^2} + \frac{1}{2!n^4} + \ldots$$ and $$\cos \left(\frac1n\right) = 1 - \frac{1}{2!n^2} + \frac{1}{4!n^4} + \ldots$$ so their difference is asymptotically equivalent to $$\frac{3}{2n^2}$$.

Can you do the denominator and compare?

The best way to solve limits like these is to split them in to standard limits.

$$\lim\limits_{n\to\infty}\frac{e^{\frac{1}{n^2}}-\cos {\frac{1}{n}}}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}$$

Hence, we can split the term in the numerator to apply standard logarithmic and trigonometric limits.

$$\lim\limits_{n\to\infty}\frac{e^{\frac{1}{n^2}}-1+1-\cos {\frac{1}{n}}}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}$$
$$\Rightarrow\lim\limits_{n\to\infty}\frac{e^{\frac{1}{n^2}}-1}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}\cdot\frac{n^2}{n^2}+\lim\limits_{n\to\infty}\frac{2\sin^2 (\frac{1}{2n})}{\frac{1}{n}\log (\frac{n+1}{n})-(\sin (\frac{2}{n}))^2}\cdot\frac{4n^2}{4n^2}$$

$$\color{blue}{\lim\limits_{h\to0}\frac{e^h-1}{h}=1}, \color{red}{\lim\limits_{h\to0}\frac{\sin h}{h}=1}$$

$$\Rightarrow\lim\limits_{n\to\infty}\color{blue}{\frac{e^{\frac{1}{n^2}}-1}{\frac{1}{n^2}}}\cdot\frac{1}{n\log(\frac{n+1}{n})-\frac{\sin^2\frac{2}{n}}{\frac{1}{n^2}}}+\lim\limits_{n\to\infty}\color{red}{\frac{2\sin^2(\frac{1}{2n})}{(\frac{1}{2n})^2}}\cdot\frac{1}{n\log(\frac{n+1}{n})-\frac{\sin^2\frac{2}{n}}{\frac{1}{n^2}}}\cdot\frac{1}{4}$$
$$\Rightarrow\lim\limits_{n\to\infty}\frac{\color{blue}{1}+\color{red}{\frac{1}{2}}}{n\log(\frac{n+1}{n})-\frac{\sin^2\frac{2}{n}}{\frac{1}{n^2}}}$$
$$\Rightarrow\lim\limits_{n\to\infty}\frac{1.5}{\frac{\log(1+\frac{1}{n})}{\frac{1}{n}}-4\frac{\sin^2(\frac{2}{n})}{\frac{4}{n^2}}}$$.

Hence the limits is $$\frac{-1}{2}$$

Hint

Set $$1/n=h$$. to find $$\dfrac{(e^{h^2}-1)/h^2+(1-\cos h)/h^2}{\ln(1+h)/h-4(\sin2h/2h)^2}$$

Use $$\lim_{x\to0}\dfrac{\sin x}x=1$$

$$\lim_{x\to0\dfrac{e^x-1}x=1=\lim_{x\to0}\dfrac{\ln(1+x}x$$