limit as t approaches infinity of sqrt(25t^2+2)-5t

解

t \to \infty lim (25 t^{2} + 2 - 5 t)

0

解答ステップ

t \to \infty lim (25 t^{2} + 2 - 5 t)

25 t^{2} + 2 - 5 t

の共役で乗じる

\frac{2}{25 t ^{2} + 2 + 5 t}

= t \to \infty lim (\frac{2}{25 t ^{2} + 2 + 5 t})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 2 \cdot t \to \infty lim (\frac{1}{25 t ^{2} + 2 + 5 t})

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

不定形式の場合を除く

= 2 \cdot \frac{lim _{t \to \infty} ( 1 )}{lim _{t \to \infty} ( 25 t ^{2} + 2 + 5 t )}

t \to \infty lim (1) = 1

t \to \infty lim (25 t^{2} + 2 + 5 t) = \infty

= 2 \cdot \frac{1}{\infty}

簡素化

2 \cdot \frac{1}{\infty} : 0

= 0