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Question Number 62067 by naka3546 last updated on 15/Jun/19

$$\underset{x\to -1}{\lim }\frac{x^3+2x^2-1}{{(x+1)}^2}=?$$

Commented by gunawan last updated on 15/Jun/19

$$\underset{x\to -1}{\lim }\frac{3x^2+4x}{2x+2}$$$$\underset{x\to -1}{\lim }\frac{6x+4}{2}=3x+2=-1$$

Commented by Tony Lin last updated on 15/Jun/19

$$\underset{x\to -1}{\lim }\frac{x^3+2x^2-1}{{(x+1)}^2}$$$$=\underset{x\to -1}{\lim }\frac{(x+1)(x^2+x-1)}{{(x+1)}^2}$$$$=\underset{x\to -1}{\lim }\frac{x^2+x-1}{x+1}\left(not\frac{0}{0}\right)$$$$\Rightarrow thevaluedoesnotexist$$

Commented by Tony Lin last updated on 15/Jun/19

$$3\times {(-1)}^2+4\times (-1)=-1\ne 0$$

Answered by MIR ZAKIR last updated on 15/Jun/19

$$lim$$$$x-\varphi$$

Answered by MJS last updated on 15/Jun/19

$$\frac{x^3+2x^2-1}{{(x+1)}^2}=\frac{x^2+x-1}{x+1}=x-\frac{1}{x+1}=$$$$\mathrm{let}x=t-1$$$$=t-1-\frac{1}{t}$$$$\underset{t\to 0}{\lim }\frac{1}{t}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}\Rightarrow \underset{t\to 0}{\lim }t-1-\frac{1}{t}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}\Rightarrow$$$$\Rightarrow \underset{x\to -1}{\lim }\frac{x^3+2x^2-1}{{(x+1)}^2}{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$

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