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I have some examples in Demidovič using this technique and there seems to be no reliable source for them online, so I'll make a small tutorial.

Example 1: $$\lim_{x\to\infty}\Bigg[\Big(\frac{2x+1}{2x+5}\Big)^{2x+3}\Bigg]$$ Example 2: $$\lim_{x\to\infty}\Bigg[\Big(\frac{x}{x+3}\Big)^x\Bigg]$$

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Okay so solutions
Preparation:
We know that: $\lim_{x\to\infty}\Big(1+\frac{1}{x}\Big)^x$ equals to $e$
We can use this fact to calculate the limits of type $1^\infty$ when $x\to\infty$
Example 1:
$$ \lim_{x\to\infty}\Bigg[\Big(\frac{2x+1}{2x+5}\Big)^{2x+3}\Bigg] $$ Prepare the internal expression:
$$ \Big(\frac{2x+1}{2x+5}\Big)^{2x+3} = \Bigg[\Big(\frac{2x+1+4}{2x+1}\Big)^{2x+3}\Bigg]^{-1} $$ Split the $\frac{2x+1+4}{2x+1}$ into $1+\frac{1}{\frac{2x+1}{4}}$. Now you need to set the exponent over the inner parenthesis to match the expression in the denominator. So your exponent needs to be $\frac{2x+1}{4}$, but you still have to retain the same value over the exponent so you normalize it by multiplying it with a number that will give $2x+3$ as a solution when multiplied. Lets call that number $b$ and calculate it $$ \frac{2x+1}{4}b = 2x+3 \implies b= \frac{8x+12}{2x+1} $$ So our expression becomes:
$$ lim_{x\to\infty}\Bigg[\Big(1+\frac{1}{\frac{2x+1}{4}}\Big)^{\frac{2x+1}{4}*\frac{8x+12}{2x+1}}\Bigg]^{-1}$$ Now we can use the first equation above and the fact that $\lim(f(x)^b)$ can be separated to get $$ \Big[e^{(\lim_{x\to\infty}\frac{8x+12}{2x+1})}\Big]^{-1} $$ Which in the end gives us $$ e^{-4}=\frac{1}{e^4} $$

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