Pergunta
lim _(xarrow +infty )(sqrt [3](7x^3+x^2)-2x);lim _(xarrow 1)((sqrt [3](5x+3-2))/(x-1));lim _(xarrow +infty )((sqrt [3](x^2+1))/(sqrt (x))) lim _(xarrow 0)[(sin(x^2))/(xtan(3x))] sqrt [4](6);sqrt [3](4);sqrt [6](13) -1 f 41121 ) f(x)=(sqrt (3x+4)-1)/(x+1);xneq -1 f(-1)=(3)/(2) ) f(x)=x^2sin((1)/(x))+1;xlt 0 f(x)=(1-sqrt (x))/(1+sqrt (x));xgeqslant 0 f(0) (vxlt 0);-x^2+1leqslant f(x)leqslant x^2+1 am lim _(xarrow 0^+)f(x) o f 2. số (xin R) ;f(x)=x^5+x^3+x-3 IRE 0lt alpha lt 1 alpha f(x)=0 0.5 alau alpha f(-alpha )=-6 f(x) (f(x))/(x^2)-4leqslant 0 R 3 (xin I=]-infty ;-1] h(x)=sqrt [3](x^2-1) lim _(xarrow -infty )h(x) , h(-1) alt bLongrightarrow h(a)gt h(b) I . h h^-1 h^-1(x)
Solução
Verification of experts
3.0310 Voting
RodrigoVeterano · Tutor por 10 anos
Responder
1. $\lim _{x\rightarrow +\infty }(\sqrt [3]{7x^{3}+x^{2}}-2x) = 0$<br /><br />2. $\lim _{x\rightarrow 1}(\frac {\sqrt [3]{5x+3-2}}{x-1}) = \infty$<br /><br />3. $\lim _{x\rightarrow +\infty }(\frac {\sqrt [3]{x^{2}+1}}{\sqrt {x}}) = 1$<br /><br />4. $\lim _{x\rightarrow 0}[\frac {sin(x^{2})}{xtan(3x)}] = 0$<br /><br />5. $\sqrt [4]{6};\sqrt [3]{4};\sqrt [6]{13}$<br /><br />6. $-1$<br /><br />7. $\{ \begin{matrix} f(x)=\frac {\sqrt {3x+4}-1}{x+1};x\neq -1\\ f(-1)=\frac {3}{2}\end{matrix} $<br /><br />8. $\{ \begin{matrix} f(x)=x^{2}sin(\frac {1}{x})+1;x\lt 0\\ f(x)=\frac {1-\sqrt {x}}{1+\sqrt {x}};x\geqslant 0\end{matrix} \}$<br /><br />9. $f(0)$<br /><br />10. $(vx\lt 0);-x^{2}+1\leqslant f(x)\leqslant x^{2}+1$<br /><br />11. $\lim _{x\rightarrow 0^{+}}f(x)$<br /><br />12. $(x\in R)\quad ;f(x)=x^{5}+x^{3}+x-3$<br /><br />13. $0\lt \alpha \lt 1$ $\alpha $ $f(x)=0$<br /><br />14. $f(-\alpha )=-6$<br /><br />15. $f(x)$<br /><br />16. $\frac {f(x)}{x^{2}-4}\leqslant 0$ $R$<br /><br />17. $(x\in I=]-\infty ;-1]\quad h(x)=\sqrt [3]{x^{2}-1}$<br /><br />18. $\lim _{x\rightarrow -\infty }h(x)$<br /><br />19. $h(-1)$<br /><br />20. $a\lt b\Longrightarrow h(a)\gt h(b)$<br /><br />21. $I. h$<br /><br />22. $h^{-1}$<br /><br />23. $h^{-1}(x)$
Clique para avaliar: