Question

Help please

Answer 1

Answer:

$\displaystyle \: \frac{5}{4}$

Step-by-step explanation:

we are given a expression

we are said to solve it using L'Hôpital's Rule

recall, L'Hôpital's Rule:

$\displaystyle \lim _{x \to \: c}( \frac{f(x)}{g(x)} ) = \lim _{x \to \: c} \frac{f ^{'}(x) }{ {g}^{'}(x) }$

it is to say the ' means derivative

our given expression:

$\displaystyle\lim_{ x\to 2}\frac{x^2+x-6}{x^2-4}$

let's apply L'Hôpital's Rule

$\displaystyle\lim_{ x\to 2}\frac{ \dfrac{d}{dx} (x^2+x-6)}{ \dfrac{d}{dx} (x^2-4)}$

some formulas of derivative

$\displaystyle \: \frac{d}{dx} {x}^{n} = {nx}^{n - 1}$

$\displaystyle \: \frac{d}{dx} {x}^{} = 1$

$\displaystyle \: \frac{d}{dx} {c}^{} = 0$

$\sf \displaystyle \: \frac{d}{dx} {f}^{} (x) + {g}^{}(x) = {f}^{'} (x) + {g}^{'}(x)$

use sum derivative formula to simplify:

$\displaystyle\lim_{ x\to 2}\frac{ \dfrac{d}{dx} (x^2)+ \dfrac{d}{dx} (x) + \dfrac{d}{dx}( -6)}{ \dfrac{d}{dx} (x^2) + \dfrac{d}{dx} (-4)}$

simplify using exponents using exponent derivative formula:

$\displaystyle\lim_{ x\to 2}\frac{ 2x+ \dfrac{d}{dx} (x) + \dfrac{d}{dx}( -6)}{ 2x + \dfrac{d}{dx} (-4)}$

use variable derivative formula to simplify variable:

$\displaystyle\lim_{ x\to 2}\frac{ 2x+ 1+ \dfrac{d}{dx}( -6)}{ 2x + \dfrac{d}{dx} (-4)}$

use constant derivative formula to simplify derivative:

$\displaystyle\lim_{ x\to 2}\frac{ 2x+ 1+ 0}{ 2x + 0}$

simplify addition:

$\displaystyle\lim_{ x\to 2}\frac{ 2x+ 1}{ 2x }$

since we are approaching x to 2

we can substitute 2 for x

$\displaystyle\lim_{ x\to 2}\frac{ 2.2+ 1}{ 2.2 }$

simplify multiplication:

$\displaystyle\frac{ 4+ 1}{ 4}$

simplify addition:

$\displaystyle \: \frac{5}{4}$

hence,

$\displaystyle\lim_{ x\to 2}\frac{ \dfrac{d}{dx} (x^2+x-6)}{ \dfrac{d}{dx} (x^2-4)} = \frac{5}{4}$

Answer 2

Answer:

5/4 is your answer hope it helps