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salve avrei bisogno del vostro aiuto con questo esercizio:

si calcoli, se esiste, attraverso l'uso dei limiti notevoli, il seguente limite:


$\lim_{x\rightarrow 0}\left ( e^{x}-1-log\left ( 1+\sqrt{\frac{x}{x+1}} \right ) \right )\cdot tan\left ( \frac{\pi }{2}-x \right )$

grazie...

Ciao, ecco lo svolgimento dettagliato:
\[\mathop {\lim }\limits_{x \to 0} \,[{e^x} - 1 - \ln (1 + \sqrt {\frac{x}{{x + 1}}} )]\;tg(\frac{\pi }{2} - x) = \]
\[ = \left\{ \begin{array}{l}
\not \exists \;\;per\;x \to {0^ - }\;\;(C.E.\;\sqrt {\frac{x}{{x + 1}}} )\\
[0 \cdot \infty ]\;\;\;per\;x \to {0^ + }
\end{array} \right.\]
calcoliamo dunque il limite solo per ${x \to {0^ + }}$
\[\mathop {\lim }\limits_{x \to {0^ + }} \,[{e^x} - 1 - \ln (1 + \sqrt {\frac{x}{{x + 1}}} )]\;tg(\frac{\pi }{2} - x) = \]
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \,[({e^x} - 1)\,tg(\frac{\pi }{2} - x) - \ln (1 + \sqrt {\frac{x}{{x + 1}}} )\,tg(\frac{\pi }{2} - x)] = \]
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \,[({e^x} - 1)\,tg(\frac{\pi }{2} - x)] - \mathop {\lim }\limits_{x \to {0^ + }} \,[\ln (1 + \sqrt {\frac{x}{{x + 1}}} )\,tg(\frac{\pi }{2} - x)] = \]
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \,[(\frac{{{e^x} - 1}}{x})x\,ctg(x)] - \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{{\ln (1 + \sqrt {\frac{x}{{x + 1}}} )}}{{\sqrt {\frac{x}{{x + 1}}} }}\sqrt {\frac{x}{{x + 1}}} \,ctg(x)] = \]
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{{{e^x} - 1}}{x}] \cdot \mathop {\lim }\limits_{x \to {0^ + }} \,[x\,ctg(x)] - \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{{\ln (1 + \sqrt {\frac{x}{{x + 1}}} )}}{{\sqrt {\frac{x}{{x + 1}}} }}] \cdot \mathop {\lim }\limits_{x \to {0^ + }} \,[\sqrt {\frac{x}{{x + 1}}} \,ctg(x)] = \]
\[ = 1 \cdot \mathop {\lim }\limits_{x \to {0^ + }} \,[x\,\frac{{\cos (x)}}{{sen(x)}}] - 1 \cdot \mathop {\lim }\limits_{x \to {0^ + }} \,[\sqrt {\frac{x}{{x + 1}}} \,\frac{{\cos (x)}}{{sen(x)}}] = \]
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{x}{{sen(x)}}\cos (x)\,] - \mathop {\lim }\limits_{x \to {0^ + }} \,[\sqrt {\frac{x}{{x + 1}}} \,\frac{{\cos (x)}}{{sen(x)}}\frac{x}{x}] = \]
\[ = \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{x}{{sen(x)}}] \cdot \mathop {\lim }\limits_{x \to {0^ + }} \,[\cos (x)\,] - \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{x}{{sen(x)}}] \cdot \mathop {\lim }\limits_{x \to {0^ + }} \,[\sqrt {\frac{x}{{x + 1}}} \,\frac{{\cos (x)}}{x}] = \]
\[ = 1 \cdot 1 - 1 \cdot \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{1}{x}\sqrt {\frac{x}{{x + 1}}} ] \cdot \mathop {\lim }\limits_{x \to {0^ + }} \,\,\cos (x) = \]
\[ = 1 - \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{1}{x}\sqrt {\frac{x}{{x + 1}}} ] \cdot 1 = 1 - \mathop {\lim }\limits_{x \to {0^ + }} \,[\frac{1}{x}\sqrt {\frac{x}{{x + 1}}} ] = \]
\[ = 1 - \mathop {\lim }\limits_{x \to {0^ + }} \,\sqrt {\frac{x}{{{x^2}(x + 1)}}} = 1 - \mathop {\lim }\limits_{x \to {0^ + }} \,\sqrt {\frac{1}{{x(x + 1)}}} = 1 - \sqrt {\frac{1}{{{0^ + }}}} = - \infty \]
Saluti.

grazie