Find the limit of (1/i) * e^(ix) as x approaches infinity.

Step-by-step solution

1. 1

   Simplify the constant $\dfrac{1}{i}$ first. Multiplying numerator and denominator by $i$ and using $i^{2}=-1$:

   $\frac{1}{i}=\frac{i}{i^{2}}=\frac{i}{-1}=-i$

   So the expression is $-ie^{ix}$. Clearing $i$ out of a denominator this way is the complex analogue of rationalising.

2. 2

   Expand with Euler formula. Since $e^{ix}=\cos x+i\sin x$,

   $-i\left(\cos x+i\sin x\right)=-i\cos x-i^{2}\sin x=\sin x-i\cos x$

   The real part is $\sin x$ and the imaginary part is $-\cos x$.

3. 3

   Test each component for a limit. A complex-valued function converges exactly when its real and imaginary parts both converge. Here $\sin x$ oscillates between $-1$ and $1$ forever and $\cos x$ does the same, so neither part has a limit as $x\to\infty$.

4. 4

   Note that the modulus does not decay. Because $\left|e^{ix}\right|=1$ and $\left|-i\right|=1$,

   $\left|\frac{1}{i}e^{ix}\right|=1\quad\text{for every }x$

   The point never spirals inward: it runs around the unit circle at constant speed, returning to where it started every $2\pi$.

5. 5

   Conclude that the limit does not exist. Two subsequences settle on different values — at $x=2\pi n$ the expression is $-i$, while at $x=\tfrac{\pi}{2}+2\pi n$ it is $1$. A limit would force these to agree, so

   $\lim_{x\to\infty}\frac{1}{i}e^{ix}\ \text{does not exist.}$