Equating the denominator to zero would be finding the vertical asymptotes, not the horizontal asymptotes. If you want to find the horizontal asymptotes, you can either use long division or use another trick:
(4x^2)/(x^2-4) = 4(x^2-4)+16/(x^2-4)
which then simplifies to 4+16/(x^2-4), meaning that the horizontal asymptote is y=4
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Equating the denominator to zero would be finding the vertical asymptotes, not the horizontal asymptotes. If you want to find the horizontal asymptotes, you can either use long division or use another trick:
(4x^2)/(x^2-4) = 4(x^2-4)+16/(x^2-4)
which then simplifies to 4+16/(x^2-4), meaning that the horizontal asymptote is y=4
lim x-->∞ 4x^2 /(x^2-4)
= lim x-->∞ 4x^2 / (x^2 (1-4/x^2)
= lim x-->∞ 4 / (1-4/x^2)
As x approaches ∞, 4/x^2 approaches 0
lim x-->∞ 4 /1 = 4
y= 4 is the horizontal asymptote
Let y = 4x²/(x² - 4)
so, y = 4(x² - 4)/(x² - 4) + 16/(x² - 4)
i.e. y = 4 + 16/(x² - 4)
As x --> ± ∞, x² - 4 --> ∞
Hence, 16/(x² - 4) --> 0
So, y --> 4 + 0....i.e. y = 4...horizontal asymptote
Well,
the limit at +/-oo of a rational fraction (quotient of two polynomials) is equivalent to the limit of the ratio of monomials of highest degree,
so, here
lim ( x ---> +/-oo ) 4x^2/(x^2 - 4) = lim ( x ---> +/-oo ) 4x^2/x^2 = 4
therefore
the horizontal line of equation y = 4 is horizontal asymptote to the curve.
hope it' ll help !!