lim-x-0-x-x-x-1-1-4-1-

Question Number 131882 by Eric002 last updated on 09/Feb/21

\lim_{x \to 0} \frac{x}{x + \sqrt[4]{x + 1} - 1}

Answered by liberty last updated on 09/Feb/21

L^{'} H \ddot{o p i t a l} L = \lim_{x \to 0} [\frac{1}{1 + \frac{1}{4 \sqrt[4]{{(x + 1)}^{3}}}}] = \frac{1}{1 + \frac{1}{4}} = \frac{4}{5}

or \lim_{x \to 0} [\frac{x}{x + 1 + \frac{x}{4} - 1}] = \lim_{x \to 0} \frac{4 x}{5 x} = \frac{4}{5}

Commented by Eric002 last updated on 09/Feb/21

h o w d i d y o u g e t \frac{x}{x + 1 + \frac{x}{4} - 1}

Commented by EDWIN88 last updated on 09/Feb/21

Bernoulli equation

\lim_{x \to 0} {(1 + f (x))}^{n} \approx \lim_{x \to 0} (1 + n . f (x))

so \lim_{x \to 0} \sqrt[4]{1 + x} \approx \lim_{x \to 0} (1 + \frac{x}{4})

Commented by Eric002 last updated on 09/Feb/21

t h a n k y o u s i r

Answered by EDWIN88 last updated on 09/Feb/21

the another way

let \sqrt[4]{x + 1} = ℓ where ℓ \to 1 and x = ℓ^{4} - 1

L = \lim_{ℓ \to 1} \frac{ℓ^{4} - 1}{ℓ^{4} + ℓ - 2} = \lim_{ℓ \to 1} \frac{(ℓ^{2} + 1) (ℓ + 1) (ℓ - 1)}{(ℓ - 1) (ℓ^{3} + ℓ^{2} + ℓ + 2)}

= \lim_{ℓ \to 1} \frac{(ℓ^{2} + 1) (ℓ + 1)}{ℓ^{3} + ℓ^{2} + ℓ + 2} = \frac{4}{5}

Answered by mathmax by abdo last updated on 10/Feb/21

let f (x) = \frac{x}{x +^{4} \sqrt{1 + x} - 1} \Rightarrow f (x) = \frac{x}{x + {(1 + x)}^{\frac{1}{4}} - 1}

\Rightarrow f (x) \sim \frac{x}{x + 1 + \frac{x}{4} - 1} = \frac{x}{\frac{5}{4} x} = \frac{4}{5} \Rightarrow \lim_{x ⌣) 0} f (x) = \frac{4}{5}