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evaluate the limit. lim 1+8x/10-3x x->inf
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8/-3
-8/3
-8/3
or if you prefer \[-\frac{8}{3}\]
if you divide by x you get \[\lim x->\infty (1/x+8)/(10/x-3)\]but the lim x->inf of a constant/x is zero so you get (0+8)/(0-3)
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\[{1+8x \over10-3x} = {x(8 + {1\over x}) \over x(-3 + {10 \over x}) } = {8 + {1\over x} \over -3 + {10 \over x} }\] \[\implies \lim_{x\rightarrow \infty}{1+8x \over10-3x} = \lim_{x\rightarrow \infty}{8 + {1\over x} \over -3 + {10 \over x} } =\ ?\]
thanks guys i appreciate the help and the step by process!
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