A question requires you to find out the greatest value for a for which a complicated limit(with variables in both the base and its exponent) equals $\frac{1}{4}$. Now, the limit simplifies down to $(\frac{sinh-ah}{h+sinh})^{1+\sqrt{1+h}}$, where h approaches 0. By direct substitution, this is $(\frac{1-a}{2})^{2}$
The answer is 0, and not 2, as the exponential function isn't defined as having a negative base, which it would if a was 2.
Apparently the limit doesn't exist due to the surrounding values of 2, and not when the exponent is exactly 2, as the expression obviously exists at that exact point and equals $\frac{1}{4}$.
So is it due to the fact that as the limit approaches the exact value, there are multiple points where the expression becomes complex?
Best Answer
The answer to the question posed in the last paragraph is, It depends on your definition of limit. It also depends on what you consider the "natural" domain of a function when the formula used to define it doesn't always make sense.
If $1-a\over2$ is negative, then $\sin h-ah\over h+\sin h$ will be negative for all values of $h$ near $0$, in which case the exponential function is rarely real (only when the exponent $1+\sqrt{1+h}$ is rational with odd denominator), so the function is undefined at most points. If your definition of limit requires the function to be defined at all points in a (punctured) neighborhood of the point being tended to, then that's enough to conclude the limit does not exist (in which case the answer to the question in the OP's last paragraph is simply Yes).
But if your definition of limit requires only that the function be defined somewhere in each (punctured) neighborhood of the point being tended to, then you have to look a bit more closely. Now the natural domain of the function $(\sin h-ah)/(h+\sin h)^{1+\sqrt{1+h}}$ (for $h$ near $0$) consists of points for which $1+\sqrt{1+h}=m/n$ with $n$ odd, so we can ask if we have a well-defined limit as $m/n\to2$.
We don't.
We don't because there are two ways to approach $2$ using fractions with odd denominator. If we approach $2$ using fractions with even numerator, e.g., ${m\over n}={2\cdot3^k+2\over3^k}$, then we indeed get $\left(1-a\over2\right)^2$. But if we approach $2$ using fractions with odd denominator, then (when $1-a\over2$ is negative) we get $-\left(1-a\over2\right)^2$. Since these values are different, the limit does not exist.
In sum, the reason the limit does not exist when $(1-a)/2$ is negative is related to the fact that the expression takes complex values at multiple points, but the precise reason is a bit more complicated, depending on how you define things.