1) Calcule o limite se existir. lim _(xarrow -2)3x^4+2x^2-x+1 lim _(xarrow 2)sqrt ((2x^2+1)/(3x-2)) lim _(xarrow -1)(x^2-4x)/(x^2)-3x-4 lim _(uarrow 0)sqrt (u^4+3u+6) d) lim _(xarrow 2)(x^2+x-6)/(x-2) lim _(tarrow -3)(t^2-9)/(2t^2)+7t+3 g) lim _(xarrow -1)(2x^2+3x+1)/(x^2)-2x-3 i) lim _(harrow 0)((h+3)^3-8)/(h) j) lim _(harrow 0)(sqrt (9+h)-3)/(h) k) lim _(xarrow -4)(frac (1)/(4)+(1)/(x))(4+x) 1) lim _(xarrow 1)2x+1 m) lim _(yarrow 6^+)(y+6)/(y^2)-36 lim _(xarrow -1)(x^2+6x+5)/(x^2)-3x-4 0) lim _(tarrow (5)/(2))(2t^2-3t-5)/(2t-5)

Vamos calcular cada limite:<br /><br />1) $\lim _{x\rightarrow -2}3x^{4}+2x^{2}-x+1 = 3(-2)^{4}+2(-2)^{2}-(-2)+1 = 48+8+2+1 = 59$<br /><br />2) $\lim _{x\rightarrow 2}\sqrt {\frac {2x^{2}+1}{3x-2}} = \sqrt {\frac {2(2)^{2}+1}{3(2)-2}} = \sqrt {\frac {9}{4}} = \frac {3}{2}$<br /><br />3) $\lim _{x\rightarrow -1}\frac {x^{2}-4x}{x^{2}-3x-4} = \frac {(-1)^{2}-4(-1)}{(-1)^{2}-3(-1)-4} = \frac {5}{2}$<br /><br />4) $\lim _{u\rightarrow 0}\sqrt {u^{4}+3u+6} = \sqrt {0^{4}+3(0)+6} = \sqrt {6}$<br /><br />5) $\lim _{x\rightarrow 2}\frac {x^{2}+x-6}{x-2} = \frac {2^{2}+2-6}{2-2} = \frac {-2}{0}$ (não existe)<br /><br />6) $\lim _{t\rightarrow -3}\frac {t^{2}-9}{2t^{2}+7t+3} = \frac {(-3)^{2}-9}{2(-3)^{2}+7(-3)+3} = \frac {0}{-12}$ (não existe)<br /><br />7) $\lim _{x\rightarrow -1}\frac {2x^{2}+3x+1}{x^{2}-2x-3} = \frac {2(-1)^{2}+3(-1)+1}{(-1)^{2}-2(-1)-3} = \frac {-2}{-1}$<br /><br />8) $\lim _{h\rightarrow 0}\frac {(h+3)^{3}-8}{h} = \frac {(0+3)^{3}-8}{0} = \frac {27-8}{0}$ (não existe)<br /><br />9) $\lim _{h\rightarrow 0}\frac {\sqrt {9+h}-3}{h} = \frac {\sqrt {9+0}-3}{0} = \frac {0}{0}$ (aplica-se a regra de L'Hôpital)<br /><br />10) $\lim _{x\rightarrow -4}\frac {\frac {1}{4}+\frac {1}{x}}{4+x} = \frac {\frac {1}{4}+\frac {1}{-4}}{4+(-4)} = \frac {-\frac {1}{4}}{0}$ (não existe)<br /><br />11) $\lim _{x\rightarrow 1}2x+1 = 2(1)+1 = 3$<br /><br />12) $\lim _{y\rightarrow 6^{+}}\frac {y+6}{y^{2}-36} = \frac {6+6}{6^{2}-36} = \frac {12}{0}$ (não existe)<br /><br />13) $\lim _{x\rightarrow -1}\frac {x^{2}+6x+5}{x^{2}-3x-4} = \frac {(-1)^{2}+6(-1)+5}{(-1)^{2}-3(-1)-4} = \frac {-4}{-2}$<br /><br />14) $\lim _{t\rightarrow \frac {5}{2}}\frac {2t^{2}-3t-5}{2t-5} = \frac {2(\frac {5}{2})^{2}-3(\frac {5}{2})-5}{2(\frac {5}{2})-5} = \frac {\frac {25}{2}-\frac {15}{2}-5}{5-5} = \frac {-\frac {5}{2}}{0}$ (não existe)