Question

\[\lim_{x \rightarrow \-3} \frac{x^5+243}{x^4-81}\]

Answer 1

\[\lim_{x \rightarrow -3 } \frac{x^5+243}{x^4-81}\]

Answer 2

\(x^5+243=x^5+3^5\)
\(=(x+3)(x^4-3x^3+9x^2-27x+81)\)
\(x^4-81=x^4-3^4\)
\(=(x^2-3^2)(x^2+3^2)\)
\(=(x-3)(x+3)(x^2+9)\)
\[\lim_{x \rightarrow -3 } \frac{x^5+243}{x^4-81}\]
\[=\lim_{x \rightarrow -3 } \frac{(x+3)(x^4-3x^3+9x^2-27x+81)}{(x-3)(x+3)(x^2+9)}\]

\[=\lim_{x \rightarrow -3 }\frac{(x^4-3x^3+9x^2-27x+81)}{(x-3)(x^2+9)}\]
Substitute x=-3 in it to find the value

Answer 3

You did a complex and lengthy way though

Answer 4

Anyone with a shorter one is still open

Answer 5

Use L'Hopital.

Answer 6

5x/4

Answer 7

5x^4/4x^3

Answer 8

= 15/4

Answer 9

where did the powers go?