I've been trying to get the limit of a function, but I don't know how.

  • The function is $\displaystyle{10^{n}\left(1 - \mathrm{e}^{\mathrm{i}t/10^{\,n}}\,\right)}$ and the solution says this converges to $-\mathrm{i}t$ as $n \to \infty$.
  • The solution also told me to make use of the Euler's formula. I have no clue how they got to $-\mathrm{i}t$.
  • Whenever I try to write $\mathrm{e}^{\mathrm{i}t/10^{n}} = \cos\left(t/10^n\right) + \mathrm{i}\,\sin\left(t/10^{n}\right)$ my function converges to $0$.

What did I do wrong ?.

Thank you!

  • $\begingroup$ You could also us L'Hospital. If you use Euler's formula, you end up with $10^{n}(1-\cos(\frac{t}{10^{n}})-i\sin(\frac{t}{10^{n}}))$ and if you now compute the limit you will finally end up with $10^{n}(-i\sin(\frac{t}{10^{n}}))$. This is not necessarily converging to zero... $\endgroup$ – Alex Jul 10 '16 at 19:36
  • $\begingroup$ Without L'Hopital: using the fact that $\frac{\sin u}{u}\xrightarrow[u\to0]{} 1$ (standard fact), this is immediate taking $u=\frac{t}{10^n}$. $\endgroup$ – Clement C. Jul 10 '16 at 19:44
  • $\begingroup$ You probably went wrong in evaluation of one of these limits $$\lim_{x\to 0}\frac{1-\cos x}{x}\\\lim_{x\to 0}\frac{\sin x}{x}$$ $\endgroup$ – user228113 Jul 10 '16 at 19:44
  • $\begingroup$ Thank you very much! I didn't think of substituting $u=\frac{t}{10^n}$ but now with your help I totally understand it! $\endgroup$ – QuestionCookie Jul 10 '16 at 20:08

As per the comment section, make the substitution $u = \frac{t}{10^n}$ so that you are computing the limit $$\lim_{u \to 0} \frac{t\left(1 - \cos u - i\sin u\right)}{u} = \lim_{u \to 0} \frac{t(1-\cos u)}{u} - \lim_{u\to 0} \frac{it \sin u}{u}$$

Now making use of the standard limits $\lim_{x\to 0}\frac{1 - \cos x}{x}= 0$ and $\lim_{x\to 0} \frac{\sin x}{x} = 1$ we have

$$\lim_{u \to 0} \frac{t(1-\cos u)}{u} - \lim_{u\to 0} \frac{it \sin u}{u} = t \cdot 0 - it \cdot 1 = -it$$


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