What is the limit as x approaches infinity of x(5^{1/x}-1) ?

The limit as x approaches infinity of x(5^{1/x}-1) is ln(5)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

limit as x approaches infinity of x(5^{1/x}-1)

x \to \infty lim (x (5^{\frac{1}{x}} - 1))

ln (5)

Decimal

1.60943 \dots

Solution steps

x \to \infty lim (x (5^{\frac{1}{x}} - 1))

Rewrite for L'Hopital

= x \to \infty lim (\frac{- 1 + 5 ^{\frac{1}{x}}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to \infty lim \frac{- \frac{l n ( 5 ) \cdot 5 ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to \infty lim \frac{- \frac{l n ( 5 ) \cdot 5 ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{l n ( 5 ) \cdot 5 ^{\frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : ln (5) \cdot 5^{\frac{1}{x}}

= x \to \infty lim (ln (5) \cdot 5^{\frac{1}{x}})

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

= ln (5) \cdot x \to \infty lim (5^{\frac{1}{x}})

Apply exponent rules

= ln (5) \cdot x \to \infty lim (e^{\frac{1}{x} l n (5)})

Apply the Limit Chain Rule:

1

= ln (5) \cdot 1

Simplify

= ln (5)

x \to 0 + lim (\frac{arctan ( x )}{x})

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

x \to 0 lim (e^{2 x} + e^{x} - e^{2 x} + x)

x \to 0 + lim (\frac{x}{e ^{- \frac{6}{x}}})

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

What is the limit as x approaches infinity of x(5^{1/x}-1) ?
The limit as x approaches infinity of x(5^{1/x}-1) is ln(5)