Question 1106383
Look for values that make the numerator equal to zero.
{{{x^3-7x^2+6x=x(x^2-7x+6)}}}
{{{x^3-7x^2+6x=x(x-1)(x-6))}}}
Potential zeros: x=0, x=1, x=6.
Verify that the denominator does not equal zero for any of these potential zeros.
If it doesn't, then it's a real zero.
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{{{lim(n->infinity,((3n^2+1)/(4n^2-8)))=lim(n->infinity,((3+1/n^2)/(4-2/n^2)))}}}
{{{lim(n->infinity,((3n^2+1)/(4n^2-8)))=(3+0)/(4-0)}}}
{{{lim(n->infinity,((3n^2+1)/(4n^2-8)))=3/4}}}