Answered

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Calculate the limitlim x => -4 [tex] \frac{x {}^{2} + 2x - 8}{x {}^{2} + 5x + 4} [/tex]

Sagot :

The limit to be calculated is:

[tex]\lim _{x\to-4}\frac{x^2+2x-8}{x^2+5x+4}[/tex]

Notice that:

[tex]\begin{gathered} \frac{x^2+2x-8}{x^2+5x+4}=\frac{(x-2)(x+4)}{(x+1)(x+4)} \\ =\frac{(x-2)}{(x+1)},x\ne-4 \end{gathered}[/tex]

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,

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

The answer is 2.

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