I'm struggling with this assignment for a couple of hours, and i dont seem to find any clues :(
I have to find the limit of this exponential function: $$\lim_{n \to \infty} (1+\frac{1}{n+2})^{2n+3} $$
I've already tried to rewrite the expression as: $$\lim_{n \to \infty} ((1+\frac{1}{n+2})^{n+2})^{\frac{2n+3}{n+2}}$$ $$\lim_{n \to \infty} \frac{2n+3}{n+2} = 2 $$ $$\lim_{n \to \infty} (1+\frac{1}{n+2})^{2n+4}$$ $$k=n+2$$ $$\lim_{n \to \infty} (1+\frac{1}{k})^{2k})$$
But it didn't help actually. I know i have to rewrite the limit to the form of: $$\lim_{n \to \infty} (1+\frac{x}{n})^{n}= e^x$$ Unfortunately i'm out of ideas now. I will really appreciate any tips from you guys. I totally stuck with this one.
You have solved the problem yourself. Just check this.
$$\lim_{n \to \infty} (1+\frac{1}{n+2})^{2n+3} $$ $$=\lim_{n \to \infty} \left(\left(1+\frac{1}{n+2}\right)^{n+2}\right)^{\frac{2n+3}{n+2}}$$ $$= \left(\lim_{(n+2) \to \infty}\left(1+\frac{1}{n+2}\right)^{n+2}\right)^{\lim_{n \to \infty}\frac{2n+3}{n+2}}$$ $$=e^{\lim_{n \to \infty} \frac{2n+3}{n+2}} =e^2 $$