what is the reasoning behind removing the limit in the exponent

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I'm studying calculus on my own. The lecture was talking about finding limits of interesting functions. Here is a concrete example.

Find the following limit: $$ \lim\limits_{x \to \infty}(1+\frac{a}{x})^x $$

Here is the solution:

Let $y=(1 + \frac{a}{x})^x$, then

$$ \ln(y) = x\ln(1 + \frac{a}{x}) $$ Now let's take a limit while $x\to\infty$ $$ \lim\limits_{x \to \infty}\ln(y)= \lim\limits_{x \to \infty} x\ln(1 + \frac{a}{x}) $$ $\ln(1 + \frac{a}{x})$ can be approximated through a Taylor series as $ \frac{a}{x} + O(\frac{1}{x^2})$. So we get the following: $$ \lim\limits_{x \to \infty}\ln(y)= \lim\limits_{x \to \infty} x(\frac{a}{x} + O(\frac{1}{x^2})) $$ which leads to the following $$ \lim\limits_{x \to \infty}\ln(y)= \lim\limits_{x \to \infty}a = a $$

Let's exponentiate both sides $$ e^{\lim\limits_{x \to \infty}\ln(y)} = e^a $$

The next step is where I'm not clear $$ \lim\limits_{x\to\infty}y=e^a $$

How did we get rid of the $\ln(y)$ and got to just the $\lim\limits_{x\to\infty}y$. I do know that $e^{\ln(b)} = b$, but what I can't figure out is how we got from $e^{\lim\limits_{x\to\infty}\ln(y)}$ to $e^{\ln(y)}$

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The exponential function is continuous, so

$$e^{\lim\limits_{x\to\infty}\ln y}=\lim_{x\to\infty}e^{\ln y}=\lim_{x\to\infty}y ;.$$