What is the limit as x approaches 0 of ((5x)^{-1}-5^{-1})/x ?

The limit as x approaches 0 of ((5x)^{-1}-5^{-1})/x is 0

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limit as x approaches 0 of ((5x)^{-1}-5^{-1})/x

x \to 0 lim (\frac{( 5 x ) ^{- 1} - 5 ^{- 1}}{x})

0

Solution steps

x \to 0 lim (\frac{( 5 x ) ^{- 1} - 5 ^{- 1}}{x})

Factor out

5 x

from

(5 x)^{- 1} - 5^{- 1} : 5 x ((5 x)^{- 1} (5 x)^{2} - (5 x)^{- 1} (5 x)^{2} \frac{5 ^{- 1}}{( 5 x ) ^{- 1}})

= x \to 0 lim \frac{5 x ( ( 5 x ) ^{- 1} ( 5 x ) ^{2} - ( 5 x ) ^{- 1} ( 5 x ) ^{2} \frac{5 ^{- 1}}{( 5 x ) ^{- 1}} )}{x}

Simplify

\frac{5 x ( ( 5 x ) ^{- 1} ( 5 x ) ^{2} - ( 5 x ) ^{- 1} ( 5 x ) ^{2} \frac{5 ^{- 1}}{( 5 x ) ^{- 1}} )}{x} : 5 (5 x - 5 x^{2})

= x \to 0 lim (5 (5 x - 5 x^{2}))

Plug in the value

x = 0

= 5 (5 \cdot 0 - 5 \cdot 0^{2})

Simplify

= 0

f (x) = \frac{2 x + 5}{3 x - 2}

\frac{\partial}{\partial y} (y^{2} e^{2 x})

\frac{d}{d x} (\frac{- π x ^{2} - 4 x ^{2} + 64 x}{8})

(x - 1) y^{^{''}} - x y^{^{'}} + y = 0

k = 1 \sum \infty \frac{5}{2 ^{k}}

What is the limit as x approaches 0 of ((5x)^{-1}-5^{-1})/x ?
The limit as x approaches 0 of ((5x)^{-1}-5^{-1})/x is 0