What is the limit as x approaches 1 of (\sqrt[4]{x}-1)/(\sqrt[5]{x)-1} ?

The limit as x approaches 1 of (\sqrt[4]{x}-1)/(\sqrt[5]{x)-1} is 5/4

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limit as x approaches 1 of (\sqrt[4]{x}-1)/(\sqrt[5]{x)-1}

\frac{5}{4}

Decimal Notation

1.25

Solution steps

Apply L'Hopital's Rule

: x \to 1 lim (\frac{\frac{1}{4 x ^{\frac{3}{4}}}}{\frac{1}{5 x ^{\frac{4}{5}}}})

= x \to 1 lim (\frac{\frac{1}{4 x ^{\frac{3}{4}}}}{\frac{1}{5 x ^{\frac{4}{5}}}})

Plug in the value

x = 1

= \frac{\frac{1}{4 \cdot 1 ^{\frac{3}{4}}}}{\frac{1}{5 \cdot 1 ^{\frac{4}{5}}}}

Simplify

\frac{\frac{1}{4 \cdot 1 ^{\frac{3}{4}}}}{\frac{1}{5 \cdot 1 ^{\frac{4}{5}}}} : \frac{5}{4}

= \frac{5}{4}

x \to 3 lim (\frac{( x ^{2 - 7 x + 12} )}{( 3 - x )})

\int \frac{y}{( y + 1 ) ^{2}} d y

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

y = \frac{1}{2 cos ^{2} ( x )}

i = 0 \sum \infty \frac{i}{2 ^{i}}

What is the limit as x approaches 1 of (\sqrt[4]{x}-1)/(\sqrt[5]{x)-1} ?
The limit as x approaches 1 of (\sqrt[4]{x}-1)/(\sqrt[5]{x)-1} is 5/4