Question

\lim_{x \to \infty}(\frac{3x}{3x+2})^{x-2}

Answer 1

$\displaystyle \lim_{x \to \infty}\left(\frac{3x}{3x+2}\right)^{x-2}$

Здесь неопределённость $\{1^{\infty}\}$ поэтому будем пользоваться вторым замечательным пределом

$\displaystyle \lim_{x \to \infty}\left(\frac{3x}{3x+2}\right)^{x-2}=\lim_{x \to \infty}\left(1-\dfrac{2}{3x+2}\right)^{(x-2)\cdot \left(-\frac{3x+2}{2}\right)\cdot \left(-\frac{2}{3x+2}\right)}=\\ \\ \\ =e^{\displaystyle \lim_{x \to \infty}-\frac{2(x-2)}{3x+2}}=e^{-2/3}$

Answer 2

To evaluate the limit:

$$\lim_{{x \to \infty}} \left(\frac{{3x}}{{3x+2}}\right)^{x-2}$$

we can rewrite it in a more manageable form. First, let's simplify the expression inside the parentheses:

$$\frac{{3x}}{{3x+2}} = \frac{{3x}}{{x(3+\frac{2}{x})}} = \frac{{3}}{{3+\frac{2}{x}}}$$

Now we have:

$$\lim_{{x \to \infty}} \left(\frac{{3}}{{3+\frac{2}{x}}}\right)^{x-2}$$

Next, let's focus on the term inside the limit:

$$\lim_{{x \to \infty}} \left(\frac{{3}}{{3+\frac{2}{x}}}\right)^{x}$$

We recognize this as the indeterminate form of $1^\infty$, which can be rewritten as:

$$\lim_{{x \to \infty}} \left(1+\frac{{3}}{{3+\frac{2}{x}}}-1\right)^{x}$$

Using the binomial expansion with $n = \frac{{3}}{{3+\frac{2}{x}}}-1$ and $a = 1$, we have:

$$\lim_{{x \to \infty}} \left(1+n\right)^{x}$$

As $x$ approaches infinity, $n$ also approaches zero. Thus, the limit becomes:

$$\lim_{{x \to \infty}} \left(1+0\right)^{x} = \lim_{{x \to \infty}} 1^{x} = 1$$

Finally, we can substitute this result back into the original expression:

$$\lim_{{x \to \infty}} \left(\frac{{3}}{{3+\frac{2}{x}}}\right)^{x-2} = 1^{x-2} = 1$$

Therefore, the limit of the given expression as $x$ approaches infinity is 1.