Question

buna seara, am de facut niste exercitii si nu ma pricep la ele as vrea zece limite si zece derivate multumesc mult :D

Answer 1

[tex]\displaystyle \lim_{x \to\pm \infty} \frac{1}{x} =0\\ \lim_{x \to 0, x\ \textgreater \ 0\ sau\ x\ \textless \ 0} \frac{1}{x} =\pm\infty\\\\ \lim_{x \to\pm \infty} (1+ \frac{1}{x} )^x=e\\ \lim_{x \to\infty} (1+ x)^{\frac{1}{x}}=e\\ \lim_{x \to\pm \infty} (1+ \frac{1}{x} )^x=e\\ \lim_{x \to\pm \infty } \frac{3x^2+4x+5}{4x^2+4} =\frac{3}{4}\\ \lim_{x \to\pm \infty } \frac{3x^2+4x+5}{4x^3+4} =0\\ \lim_{x \to\pm \infty } \frac{3x^3+4x+5}{4x^2+4} =\pm \infty \\[/tex]
[tex]\displaystyle \lim_{x \to0,x\ \textgreater \ 0 } lnx=-\infty\\ \lim_{x \to+\infty } \sqrt{x+1}- \sqrt{x} =0[/tex]
$\displaystyle \lim_{x \to+ \infty} \frac{1}{lnx} =0$

[tex]constanta'=0\\ x'=1\\ (x^{n})'=nx^{n-1}\\ (e^x)'=e^x\\ (a^x)'=a^x\cdot lna\\ ( \sqrt{x} )'= \frac{1}{2 \sqrt{x} } \\ (lnx)'= \frac{1}{x} \\ (sinx)'=cosx\\ (cosx)'=-sinx\\ (arctgx)'= \frac{1}{x^2+1} \\[/tex]