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(x^3 + ax^2 + x + b )/ (x^2 - 4) a,b = ? if lim(f(x))\Rightarrow 5 with x\Rightarrow 2

difficulty advanced
 Aug 29, 2015

Best Answer 

 #1
avatar+118608 
+5

 

(x^3 + x^2*a + x + b)/ x^2 - 4 

 

$$if \;\;lim(f(x))\Rightarrow 5\;\; with \;\;x\Rightarrow 2$$

 

This is what I think :/

 

$$\\\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x^2-2}=5\\\\
\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x^2-2}*(x-2)=\displaystyle\lim_{x\rightarrow2}\;5(x-2)\\\\
\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x+2}=\displaystyle\lim_{x\rightarrow2}\;5x-10\\\\
\displaystyle\lim_{x\rightarrow2}\;x^3+ax^2+x+b=\displaystyle\lim_{x\rightarrow2}\;(5x-10)(x+2)\\\\
8+4a+2+b=0\\\\
4a+b=-10$$

 

 Aug 29, 2015
 #1
avatar+118608 
+5
Best Answer

 

(x^3 + x^2*a + x + b)/ x^2 - 4 

 

$$if \;\;lim(f(x))\Rightarrow 5\;\; with \;\;x\Rightarrow 2$$

 

This is what I think :/

 

$$\\\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x^2-2}=5\\\\
\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x^2-2}*(x-2)=\displaystyle\lim_{x\rightarrow2}\;5(x-2)\\\\
\displaystyle\lim_{x\rightarrow2}\;\frac{x^3+ax^2+x+b}{x+2}=\displaystyle\lim_{x\rightarrow2}\;5x-10\\\\
\displaystyle\lim_{x\rightarrow2}\;x^3+ax^2+x+b=\displaystyle\lim_{x\rightarrow2}\;(5x-10)(x+2)\\\\
8+4a+2+b=0\\\\
4a+b=-10$$

 

Melody Aug 29, 2015

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