What is the limit as x approaches 4 of (x^2-3x-4)/(x^2-8x+16) ?

The limit as x approaches 4 of (x^2-3x-4)/(x^2-8x+16) is diverges

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limit as x approaches 4 of (x^2-3x-4)/(x^2-8x+16)

x \to 4 lim (\frac{x ^{2} - 3 x - 4}{x ^{2} - 8 x + 16})

diverges

Solution steps

x \to 4 lim (\frac{x ^{2} - 3 x - 4}{x ^{2} - 8 x + 16})

x \to a - lim f (x) \neq = x \to a + lim f (x)

then the limit does not exist

x \to 4 + lim (\frac{x ^{2} - 3 x - 4}{x ^{2} - 8 x + 16}) = \infty

x \to 4 - lim (\frac{x ^{2} - 3 x - 4}{x ^{2} - 8 x + 16}) = - \infty

= diverges

x \to 7 + lim (\frac{x ^{2} - 4 x - 21}{∣ x - 7 ∣})

x \to 5 - lim (arcsin (\frac{x}{5}))

x \to 0 lim ((1 + x)^{\frac{1}{x}} - e)

x \to - \infty lim (2 \cdot (x))

x \to 0 lim (\frac{1}{3 x ^{\frac{2}{3}}})

What is the limit as x approaches 4 of (x^2-3x-4)/(x^2-8x+16) ?
The limit as x approaches 4 of (x^2-3x-4)/(x^2-8x+16) is diverges