Question

Limit x tends to -2 x^7+128/x^3+8/

Answer 1

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Answer 2

The answer is $\frac{112}{3}$

The given is $\lim_{x \to -2} (\frac{x^7+128}{x^3+8} )$

Find the value of the given equation

Solution-In mathematics, more specifically calculus, L'Hôpital's rule or L'Hospital's rule, also known as Bernoulli's rule, is a theorem which provides a technique to evaluate limits of indeterminate forms.

$\lim_{x \to -2} (\frac{x^7+128}{x^3+8} )$

Apply L'Hospital Rule

So, $\lim_{x \to -2} (\frac{7x^6}{3x^2} )$

$=\frac{7x^4}{3}$

$\lim_{x \to -2} \frac{7\times 16}{3} \\=\frac{112}{3}$

So, the value is $\frac{112}{3}$

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