Jawab:a)
[tex]\lim_{x\to \-2} 3x^{2} -5x^{2} +5x-3\\= 3(2)^{2} -5(2)^{2} +5(2)-3\\= 12-20+10-3\\=-1[/tex]
b)[tex]\lim_{x \to \-3} (3x^{2} )(x^{2} -1)\\=(3(3)^2)((3)^2-1)\\= (27)(8)\\=216[/tex]
c)
[tex]\lim_{x\to \-1} \frac{x+9}{x+6} \\=\frac{1+9}{1+6}\\ =\frac{10}{7}[/tex]
d)
[tex]\lim_{x \to \-4} \sqrt{2x+8}\\=\sqrt{2(4)+8} \\=\sqrt{8+8}\\= \sqrt{16}\\=4[/tex]
e)
[tex]\lim_{x \to \-2} (4+x^{2} )(x-3)\\ =(4+2^{2})(2-3)\\ =(4+4)(-1)\\= (8)(-1)\\=-8[/tex]
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Jawab:
a)
[tex]\lim_{x\to \-2} 3x^{2} -5x^{2} +5x-3\\= 3(2)^{2} -5(2)^{2} +5(2)-3\\= 12-20+10-3\\=-1[/tex]
b)
[tex]\lim_{x \to \-3} (3x^{2} )(x^{2} -1)\\=(3(3)^2)((3)^2-1)\\= (27)(8)\\=216[/tex]
c)
[tex]\lim_{x\to \-1} \frac{x+9}{x+6} \\=\frac{1+9}{1+6}\\ =\frac{10}{7}[/tex]
d)
[tex]\lim_{x \to \-4} \sqrt{2x+8}\\=\sqrt{2(4)+8} \\=\sqrt{8+8}\\= \sqrt{16}\\=4[/tex]
e)
[tex]\lim_{x \to \-2} (4+x^{2} )(x-3)\\ =(4+2^{2})(2-3)\\ =(4+4)(-1)\\= (8)(-1)\\=-8[/tex]