I don't understand french

Let $N(x) = \sqrt{x^2+2x+3} - \sqrt{x^2+3}$.
This is an indeterminate form of type $\infty - \infty$. We will multiply by the conjugate expression:

$$N(x) = \frac{\left(\sqrt{x^2+2x+3} - \sqrt{x^2+3}\right)\left(\sqrt{x^2+2x+3} + \sqrt{x^2+3}\right)}{\sqrt{x^2+2x+3} + \sqrt{x^2+3}}$$ $$N(x) = \frac{(x^2+2x+3) - (x^2+3)}{\sqrt{x^2+2x+3} + \sqrt{x^2+3}}$$ $$N(x) = \frac{2x}{\sqrt{x^2+2x+3} + \sqrt{x^2+3}}$$

To evaluate this limit as $x \to \infty$, we divide the numerator and the denominator by $x$ (noting that $\sqrt{x^2} = x$ for $x > 0$):

$$N(x) = \frac{2}{\sqrt{\frac{x^2+2x+3}{x^2}} + \sqrt{\frac{x^2+3}{x^2}}}$$ $$N(x) = \frac{2}{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}} + \sqrt{1+\frac{3}{x^2}}}$$

Now, we can evaluate the limit as $x \to \infty$:

$$\lim_{x \to \infty} N(x) = \frac{2}{\sqrt{1+0+0} + \sqrt{1+0}} = \frac{2}{1+1} = 1$$

Let $D(x) = 1 - \cos\left(\sin^{-1}\left(\tan\left(\frac{1}{x}\right)\right)\right)$.
Let $u = \frac{1}{x}$. As $x \to \infty$, $u \to 0^+$.
The denominator becomes:

$$D(u) = 1 - \cos\left(\sin^{-1}\left(\tan(u)\right)\right)$$

As $u \to 0^+$:

• $\tan(u) \to \tan(0) = 0$.
• $\sin^{-1}(\tan(u)) \to \sin^{-1}(0) = 0$.
• $\cos(\sin^{-1}(\tan(u))) \to \cos(0) = 1$. So, $\lim_{u \to 0^+} D(u) = 1 - 1 = 0$.

We use equivalent functions for $u \to 0^+$:

• $\tan(u) \sim u$
• $\sin^{-1}(v) \sim v$ (for $v \to 0$) So, $\sin^{-1}(\tan(u)) \sim \tan(u) \sim u$. Let $w = \sin^{-1}(\tan(u))$. Then $w \sim u$ as $u \to 0^+$.
• $1 - \cos(w) \sim \frac{w^2}{2}$ (for $w \to 0$) Substituting $w \sim u$, we get: $D(u) \sim \frac{u^2}{2}$. Since $u = \frac{1}{x}$, we have $D(x) \sim \frac{(1/x)^2}{2} = \frac{1}{2x^2}$. As $x \to \infty$, $\frac{1}{2x^2}$ is always positive. Therefore, the denominator approaches 0 from the positive side ($0^+$).

The original limit is $\lim_{x \to \infty} \frac{N(x)}{D(x)}$.
We found $\lim_{x \to \infty} N(x) = 1$.
We found $\lim_{x \to \infty} D(x) = 0^+$.
Therefore, the limit is of the form $\frac{1}{0^+}$, which tends to positive infinity.

$$\lim_{x \to \infty} \frac{\sqrt{x^2+2x+3} - \sqrt{x^2+3}}{1 - \cos\left(\sin^{-1}\left(\tan\left(\frac{1}{x}\right)\right)\right)} = \frac{1}{0^+} = \infty$$

The final answer is $\infty$.