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30 votes
Calculate the limitlim x => -4
\frac{x {}^(2) + 2x - 8}{x {}^(2) + 5x + 4}

User Gonan
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1 Answer

16 votes
16 votes

The limit to be calculated is:


\lim _(x\to-4)(x^2+2x-8)/(x^2+5x+4)

Notice that:


\begin{gathered} (x^2+2x-8)/(x^2+5x+4)=((x-2)(x+4))/((x+1)(x+4)) \\ =((x-2))/((x+1)),x\\e-4 \end{gathered}

Remember that, in the limit when x->-4, the value of x approaches to -4, but it never is -4. Thus, we can use the last line of the identity above,


\begin{gathered} \Rightarrow\lim _(x\to-4)(x^2+2x-8)/(x^2+5x+4)=\lim _(x\to-4)((x-2)(x+4))/((x+1)(x+4))=\lim _(x\to-4)((x-2))/((x+1))=((-4-2))/((-4+1))=-(6)/(-3)=2 \\ \Rightarrow\lim _(x\to-4)(x^2+2x-8)/(x^2+5x+4)=2 \end{gathered}

The answer is 2.

User Mike Precup
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