We have, lim_x→ 3x^5 243x^2 9 This is 00 form.So, apply L hospital rule lim_x→ 35x^4 02x 0 lim_x→ 35x^42x lim_x→ 35x^32 =5× 3^32 ...

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Algebra of Limits

limx → 3x5-24...

Question

$l i m_{x \to 3} \frac{x^{5} - 243}{x^{2} - 9}$

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Solution

We have,
${lim}_{x \to 3} \frac{x^{5} - 243}{x^{2} - 9}$

This is $\frac{0}{0}$ form.
So, apply L-hospital rule
${lim}_{x \to 3} \frac{5 x^{4} - 0}{2 x - 0}$
${lim}_{x \to 3} \frac{5 x^{4}}{2 x}$
${lim}_{x \to 3} \frac{5 x^{3}}{2}$
$= \frac{5 \times 3^{3}}{2}$
$= \frac{135}{2}$

Hence, this is the answer.

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