Question

Evaluate the given limit plz solve this.​

Answer 1

Answer:

I think x⁴+128÷8=x⁴+16.

step by step explanation:- x⁷+128÷x³+8

= x⁷-³ (x⁷÷x³=x⁷-³)

=x⁴+128÷8

=x⁴+16.

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Answer 2

To Evaluate :-

• $\sf\lim\limits_{x\to(-2)}\left[ \dfrac{x^7+128}{x^3+8}\right]$

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Solution :-

• $\sf\lim\limits_{x\to(-2)}\left[ \dfrac{x^7+128}{x^3+8}\right]$

By directly substituting value of x = -2, we get :

$\sf\to\left[ \dfrac{( - 2)^7+128}{( - 2)^3+8}\right] = \left[ \dfrac{ - 128 + 128}{ - 8 + 8} \right] = \dfrac{0}{0}$

This is an indeterminate quantity, hence substitution method fails.

Now let's apply L'Hopital rule according to which,

$\sf\mapsto \red{\lim\limits_{x\to a }\dfrac{f(x)}{g(x)} = \lim\limits_{x\to a } \dfrac{f'(x)}{g('x)}}$

So,

$\sf\implies\lim\limits_{x\to(-2)}\left[ \dfrac{x^7+128}{x^3+8}\right]$

$\sf\implies\lim\limits_{x\to(-2)}\left[ \dfrac{\frac{d}{dx}(x^7+128)}{\frac{d}{dx}(x^3+8)}\right]$

$\sf \implies\lim\limits_{x\to(-2)}\left[ \dfrac{ \frac{d}{dx}(x^7) + \frac{d}{dx}(128)}{\frac{d}{dx}(x^3)+ \frac{d}{dx} (8)}\right]$

$\sf \implies\lim\limits_{x\to(-2)}\left[ \dfrac{ 7x^6+ 0}{3x^2+ 0}\right]$

$\sf \implies\lim\limits_{x\to(-2)}\left[ \dfrac{ 7x^6}{3x^2}\right]$

$\sf \implies\lim\limits_{x\to(-2)}\left[ \dfrac{ 7x^4}{3}\right]$

Now by substituting x = -2 , we get :

$\sf \leadsto\left[ \dfrac{ 7( - 2)^4}{3}\right]$

$\sf \leadsto\left[ \dfrac{ 7 \times 16}{3}\right]$

$\sf \leadsto\left[ \dfrac{ 112}{3}\right]$

$\sf \: {Hence, \: \underline{ \underline{\lim\limits_{x\to(-2)}\left[ \dfrac{x^7+128}{x^3+8}\right] = \dfrac{112}{3} }}}$