Determiner-lim-x-3-x-3-3-x-5-2-

Question Number 198123 by a.lgnaoui last updated on 10/Oct/23

Determiner

\lim_{x \to 3} \frac{x - 3}{^{3} \sqrt{x + 5} - 2}

Answered by Mathspace last updated on 10/Oct/23

a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2}) \Rightarrow

a - b = (^{3} \sqrt{a} -^{3} \sqrt{b}) ({(^{3} \sqrt{a})}^{2} +^{3} \sqrt{a b} + {(^{3} \sqrt{b})}^{2}) \Rightarrow

(^{3} \sqrt{x + 5}) - (^{3} \sqrt{8})

= \frac{x + 5 - 8}{{(x + 5)}^{\frac{2}{3}} + {(8 (x + 5))}^{\frac{1}{3}} + 8^{\frac{2}{3}}}

\Rightarrow {l i m}_{x \to 3} f (x)

= {l i m}_{x \to 3} \frac{x - 3}{\frac{x - 3}{{(x + 5)}^{\frac{2}{3}} + 2 {(x + 5)}^{\frac{1}{3}} + 4}}

= {l i m}_{x \to 3} {(x + 5)}^{\frac{2}{3}} + 2 {(x + 5)}^{\frac{1}{3}} + 4

= 8^{\frac{2}{3}} + 2 \times 8^{\frac{1}{3}} + 4

= 4 + 4 + 4 = 12

Commented by a.lgnaoui last updated on 10/Oct/23

exact

Answered by MM42 last updated on 10/Oct/23

\sqrt[3]{x + 5} - 2 = u \Rightarrow x + 5 = {(u + 2)}^{3} \Rightarrow x = {(u + 2)}^{3} - 5

\Rightarrow {l i m}_{u \to 0} \frac{{(u + 2)}^{3} - 8}{u} = {l i m}_{u \to 0} \frac{u (u^{2} + 6 u + 12)}{u} = 12 ✓

Answered by cortano12 last updated on 11/Oct/23

\underset{―}{\underset{⏟}{}} \underset{⏟}{Y} 3