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Help with the following limit:

$$\lim_{x\to -1} \frac{x^4-3x^2-x+1}{x^3+1}$$

Plz solve step by step.

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    You have to say what you tried. – 2017-01-28
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    Please add more context to the question, such as what you tried. – 2017-01-28
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    Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. What have you attempted? Where are you stuck? – 2017-01-28
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    Hint: if you plug in $x=-1$, you get $0/0$, which means that you should be able to factor $(x+1)$ out of the numerator and denominator. – 2017-01-28

3 Answers 3

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When you factor the numerator you get $$(x+1)(x^3-x^2-2x+1)$$ When you factor the denominator you get $$(x+1)(x^2-x+1)$$

From then on, you can just check for $x=-1$.

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$$\lim _{t\to 0}\left(\frac{\left(t-1\right)^4-3\left(t-1\right)^2-\left(t-1\right)+1}{\left(t-1\right)^3+1}\right) =\lim _{t\to \:0}\left(\frac{t^3+3t-4t^2+1}{t^2-3t+3}\right)=\color{red}{\frac{1}{3}}$$

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You can use L'Hopital's rule: $$\lim_{x \to -1} \frac{x^4-3x^2-x+1}{x^3+1}=\lim_{x \to -1} \frac{\frac{d}{dx}(x^4-3x^2-x+1)}{\frac{d}{dx}(x^3+1)}$$ Evaluating the derivatives gives: $$\lim_{x \to -1} \frac{4x^3-6x-1}{3x^2}=\frac{1}{3}$$

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    Your method is certainly the fastest, for this type of limit – 2017-01-28