Question

Предел, стремящийся к бесконечности 2x[ln(x+3)-ln(x-3)] (решить,не пользуясь правилом Лопиталя)

Answer 1

$\lim\limits _{x \to \infty}\, 2x\Big (ln(x+3)-ln(x-3)\Big )=\lim\limits _{x \to \infty}ln\Big (\frac{x+3}{x-3}\Big )^{2x}=\\\\=ln\, \lim\limits _{x \to \infty}\Big (\frac{x+3}{x-3}\Big )^{2x}=ln\lim\limits _{x \to \infty}\Big (\Big (1+\frac{6}{x-3}\Big )^{\frac{x-3}{6}}\Big )^{\frac{6\cdot 2x}{x-3}}=\\\\=ln\, e^{\lim\limits _{x \to \infty}\frac{12x}{x-3}}=ln\, e^{12}=12$

Answer 2

To find the limit of the expression as x approaches infinity without using L'Hôpital's rule, we can simplify the expression first. The given expression is:

Lim[x -> ∞] 2x[ln(x+3) - ln(x-3)]

Step 1: Apply the properties of logarithms

ln(a) - ln(b) = ln(a/b)

Lim[x -> ∞] 2x[ln((x+3)/(x-3))]

Step 2: Simplify the argument of the logarithm

As x approaches infinity, (x+3)/(x-3) will also approach infinity, and we can treat it as x/x = 1.

Lim[x -> ∞] 2x[ln(1)]

Step 3: ln(1) is equal to 0.

Lim[x -> ∞] 2x * 0

Any value multiplied by 0 is equal to 0.

Therefore, the limit of the expression as x approaches infinity is:

Lim[x -> ∞] 2x[ln(x+3) - ln(x-3)] = 0.