1
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.