limit as x approaches 2 of 1/(x^2-4x+4)

解答

x \to 2 lim (\frac{1}{x ^{2} - 4 x + 4})

\infty

求解步骤

x \to 2 lim (\frac{1}{x ^{2} - 4 x + 4})

如果

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

则

x \to a lim f (x) = L

x \to 2 - lim (\frac{1}{x ^{2} - 4 x + 4}) = \infty

x \to 2 + lim (\frac{1}{x ^{2} - 4 x + 4}) = \infty

= \infty

in v erse l a pl a ce \frac{5}{s} - \frac{5}{s} e^{- 5 s}

n = 0 \sum \infty - (- 1)^{n + 1}

\frac{d}{d x} (\frac{3}{4 x ^{2}})

t an g e n t f (x) = x^{4} - 13 x^{2} + 36

\int_{0}^{1} \frac{45}{x ^{5}} d x