help please
simplify the following using L'Hôpital's Rule:
[tex]\displaystyle\lim_{x\to 2}\frac{x^2+x-6}{x^2-4}[/tex]
[tex]\text{note:explanation is a must}[/tex]
Answers
Answer:
[tex] \displaystyle \: \frac{5}{4} [/tex]
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:
[tex] \displaystyle \lim _{x \to \: c}( \frac{f(x)}{g(x)} ) = \lim _{x \to \: c} \frac{f ^{'}(x) }{ {g}^{'}(x) } [/tex]
it is to say the ' means derivative
our given expression:
[tex]\displaystyle\lim_{ x\to 2}\frac{x^2+x-6}{x^2-4}[/tex]
let's apply L'Hôpital's Rule
[tex]\displaystyle\lim_{ x\to 2}\frac{ \dfrac{d}{dx} (x^2+x-6)}{ \dfrac{d}{dx} (x^2-4)}[/tex]
some formulas of derivative
[tex] \displaystyle \: \frac{d}{dx} {x}^{n} = {nx}^{n - 1} [/tex]
[tex] \displaystyle \: \frac{d}{dx} {x}^{} = 1[/tex]
[tex] \displaystyle \: \frac{d}{dx} {c}^{} = 0[/tex]
[tex] \sf \displaystyle \: \frac{d}{dx} {f}^{} (x) + {g}^{}(x) = {f}^{'} (x) + {g}^{'}(x)[/tex]
use sum derivative formula to simplify:
[tex]\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)}[/tex]
simplify using exponents using exponent derivative formula:
[tex]\displaystyle\lim_{ x\to 2}\frac{ 2x+ \dfrac{d}{dx} (x) + \dfrac{d}{dx}( -6)}{ 2x + \dfrac{d}{dx} (-4)}[/tex]
use variable derivative formula to simplify variable:
[tex]\displaystyle\lim_{ x\to 2}\frac{ 2x+ 1+ \dfrac{d}{dx}( -6)}{ 2x + \dfrac{d}{dx} (-4)}[/tex]
use constant derivative formula to simplify derivative:
[tex]\displaystyle\lim_{ x\to 2}\frac{ 2x+ 1+ 0}{ 2x + 0}[/tex]
simplify addition:
[tex]\displaystyle\lim_{ x\to 2}\frac{ 2x+ 1}{ 2x }[/tex]
since we are approaching x to 2
we can substitute 2 for x
[tex]\displaystyle\lim_{ x\to 2}\frac{ 2.2+ 1}{ 2.2 }[/tex]
simplify multiplication:
[tex]\displaystyle\frac{ 4+ 1}{ 4}[/tex]
simplify addition:
[tex] \displaystyle \: \frac{5}{4} [/tex]
hence,
[tex]\displaystyle\lim_{ x\to 2}\frac{ \dfrac{d}{dx} (x^2+x-6)}{ \dfrac{d}{dx} (x^2-4)} = \frac{5}{4} [/tex]
Answer:
5/4 is your answer hope it helps
![Help Pleasesimplify The Following Using L'Hpital's Rule:[tex]\displaystyle\lim_{x\to 2}\frac{x^2+x-6}{x^2-4}[/tex][tex]\text{note:explanation](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/PexHhr1GyXQ9gvBMSw94ktDYD95eLhOB.jpeg)
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Answers
Answer:
Step-by-step explanation:
From points (-5,0),(1,0),(4,0) we know it has the form
y=(x+5)(x-1)(x-4)
y=(x^2+4x-5)(x-4)
y=x^3-4x^2+4x^2-16x-5x+20
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Mr. Carlevato and Mr. Burrows get into a debate about two different expressions. · Mr. Carlevato says "2(2x + 3) is equal to 4x + 6". · Mr. Burrows says "Nope, wrong, those are clearly different expressions" Try substituting x=5 into both expressions. Do you get the same answer for both expressions? Show your work!
Answers
Answer:
Yes, the same answer is obtained.
Step-by-step explanation:
According to Mr. Carlevato, [tex]2(2x+3)[/tex] is equal to [tex]4x+6[/tex]. However, [tex]2(2x+3)[/tex] is not equal to [tex]4x+6[/tex] according to Mr. Burrows.
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Step-by-step explanation:
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Step-by-step explanation:
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Step-by-step explanation:
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