Question

limit x tends to -3 : x³ +27 ÷ x power 5 +243

Answer 1

$\displaystyle\mathsf{\lim_{x\to\,-3}\;\frac{x^3+27}{x^5+243}}$

$=\displaystyle\mathsf{\lim_{x\to\,-3}\;\frac{x^3-(-27)}{x^5-(-243)}}$

$=\displaystyle\mathsf{\lim_{x\to\,-3}\;\frac{x^3-(-3)^3}{x^5-(-3)^5}}$

$\textsf{using}$

$\boxed{\mathsf{a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+...+ab^{n-2}+b^{n-1})}}$

$=\displaystyle\mathsf{\lim_{x\to\,-3}\;\frac{(x-(-3))(x^2-3x+9)}{(x-(-3))(x^4+x^3(-3)+x^2(-3)^2+x(-3)^3+(-3)^4)}}$

$=\displaystyle\mathsf{\lim_{x\to\,-3}\;\frac{(x+3))(x^2-3x+9)}{(x+3)(x^4-3x^3+9x^2-27x+81)}}$

$=\displaystyle\mathsf{\lim_{x\to\,-3}\;\frac{x^2-3x+9}{x^4-3x^3+9x^2-27x+81}}$

$=\displaystyle\mathsf{\frac{(-3)^2-3(-3)+9}{(-3)^4-3(-3)^3+9(-3)^2-27(-3)+81}}$

$=\displaystyle\mathsf{\frac{9+9+9}{81+81+81+81+81}}$

$=\displaystyle\mathsf{\frac{27}{5{\times}81}}$

$=\displaystyle\mathsf{\frac{1}{5{\times}3}}$

$=\displaystyle\mathsf{\frac{1}{15}}$

$\implies\boxed{\mathsf{\lim_{x\to\,-3}\;\frac{x^3+27}{x^5+243}}=\frac{1}{15}}$

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