What's wrong in the evaluation of the limit of the sequence $$a_{n+1} = 4a_n+3$$ when $a_0$(first term) $=1$

Taking limits to infinty on both sides :

$\lim_{n \rightarrow \infty}a_{n+1} = 4 \lim_{n \rightarrow \infty}a_n + 3$

But $\lim_{n \rightarrow \infty}a_{n+1} =\lim_{n \rightarrow \infty}a_n$

So we get limit as $-1$. However this limit can't be negative as the sequence is increasing. Can someone point out the error with an example of a correct evaluation technique?

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    $\begingroup$ You are taking limit of $(a_n)$ and doing arithmetics with it without knowing it exists in $\mathbb{R}$. $\endgroup$ Jun 18 at 14:29
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    $\begingroup$ ...and, in fact and as continuation of the first comment, what you've proved is that the limit of that sequence, even if it exists (and it does in a generalized sense), cannot be a finite one. $\endgroup$
    – DonAntonio
    Jun 18 at 14:56
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    $\begingroup$ As a side note: You have also discovered that $-1$ is a fixed point of your transformation, in other words, if you were to start for instance with $a_1=-1$, you would have $a_n=-1$ for all $n$. $\endgroup$ Jun 18 at 15:12

Indeed, we have $a_{n+1}+1=4(a_n+1)$, which means that the sequence $\{a_n+1\}_{n\geq 0}$ is a geometric series, therefore, $$a_n+1=2\cdot 4^{n}.$$ So $$a_n=2\cdot 4^{n}-1.$$ Obviously $\lim\limits_{n\to\infty}a_n=+\infty$, so we can't take limits as you have done above.


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