# How to solve indeterminations of the type $0/0$

I am unable to find these limits:

1) $$\lim_{x \to 1} \frac{3(1 - x^2) - 2(1 - x^3)}{(1 - x^3)(1 - x^2)}$$

2) $$\lim_{x \to 0} \frac{\sqrt{1 - 2x - x^2} - (x + 1)}{x}$$

3) $$\lim_{x \to 0} \frac{\sqrt{x + 2} + \sqrt{x + 6} - \sqrt{6} - \sqrt{2}}{x}$$

4) $$\lim_{x \to 0} \frac{1 - \sqrt[3]{1 - x}}{1 + \sqrt[3]{3x - 1}}$$

My interest is not in the answers, but in the algebraic manipulations i can use to eliminate the indeterminations of the type $0/0$.

My english skills are not so good, i'm sorry for this.

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It doesn't look like these will be fixable by standard algebraic manipulations - you should use L'Hopital's rule (en.wikipedia.org/wiki/L%27Hopital%27s_rule). – Matthew Pressland Dec 14 '12 at 17:07
The top one at least can be factored. – user45150 Dec 14 '12 at 17:13
These are too many. Please choose one of them. – Amr Dec 14 '12 at 17:13
@Amr I disagree. If they are related to have the same general approach, then it makes sense to have all these questions in one problem, so responders know what techniques to be supplying – user45150 Dec 14 '12 at 17:15
@user45150 Yes but if they can be solved by a similar technique then it suffices to show how one of them is done. – Amr Dec 14 '12 at 17:17

• For 1) Notice that $x=-1$ is a zero of the numerator and denominator polynomials. So we can divide both terms by $x-1$, e.g. using polynomial long division. We get $$\begin{equation*} \frac{3(1-x^{2})-2(1-x^{3})}{x-1}=\frac{2x^{3}-3x^{2}+1}{x-1}=2x^{2}-x-1 \end{equation*}$$ $$\begin{equation*} \frac{(1-x^{3})(1-x^{2})}{x-1}=\frac{x^{5}-x^{3}-x^{2}+1}{x-1} =x^{4}+x^{3}-x-1,\end{equation*}$$ both of which can be divided again by $x-1$ $$\begin{eqnarray*} \frac{2x^{2}-x-1}{x-1} &=&2x+1 \\ \frac{x^{4}+x^{3}-x-1}{x-1} &=&x^{3}+2x^{2}+2x+1. \end{eqnarray*}$$ We thus have $$\begin{equation*} \frac{3(1-x^{2})-2(1-x^{3})}{(1-x^{3})(1-x^{2})}=\frac{2x^{2}-x-1}{ x^{4}+x^{3}-x-1}=\frac{2x+1}{x^{3}+2x^{2}+2x+1}. \end{equation*}$$ So $$\begin{equation*} \lim_{x\rightarrow 1}\frac{3(1-x^{2})-2(1-x^{3})}{(1-x^{3})(1-x^{2})} =\lim_{x\rightarrow 1}\frac{2x+1}{x^{3}+2x^{2}+2x+1}=\frac{1}{2}. \end{equation*}$$ Alternatively we can use the identities $$\begin{eqnarray*} 1-x^{2} &=&(1-x)(1+x) \\ 1-x^{3} &=&(1-x)\left( x^{2}+x+1\right) \end{eqnarray*}$$ to factor both terms as follows: $$\begin{eqnarray*} \frac{3(1-x^{2})-2(1-x^{3})}{(1-x^{3})(1-x^{2})} &=&\frac{ 3(1-x)(1+x)-2(1-x)\left( x^{2}+x+1\right) }{(1-x)\left( x^{2}+x+1\right) (1-x)(1+x)} \\ &=&\frac{3(1+x)-2\left( x^{2}+x+1\right) }{\left( x^{2}+x+1\right) (1-x)(1+x) } \\ &=&\frac{-2x^{2}+x+1}{\left( x^{2}+x+1\right) (1-x)(1+x)} \\ &=&\frac{-2(x+\frac{1}{2})(x-1)}{\left( x^{2}+x+1\right) (1-x)(1+x)} \\ &=&\frac{2(x+\frac{1}{2})}{\left( x^{2}+x+1\right) (1+x)}. \end{eqnarray*}$$
• For 2) Expand the fraction and multiply the new fraction by the conjugate of the numerator $$\begin{eqnarray*} &&\frac{\sqrt{1-2x-x^{2}}-(x+1)}{x} \\ &=&-1+\frac{\sqrt{1-2x-x^{2}}-1}{x} \\ &=&-1+\frac{\left( \sqrt{1-2x-x^{2}}-1\right) \left( \sqrt{1-2x-x^{2}} +1\right) }{x\left( \sqrt{1-2x-x^{2}}+1\right) } \\ &=&-1+\frac{1-2x-x^{2}-1}{x\left( \sqrt{1-2x-x^{2}}+1\right) }=-1-\frac{2+x}{ \sqrt{1-2x-x^{2}}+1}. \end{eqnarray*}$$ Hence $$\begin{equation*} \lim_{x\rightarrow 0}\frac{\sqrt{1-2x-x^{2}}-(x+1)}{x}=-1-\lim_{x\rightarrow 0}\frac{2+x}{\sqrt{1-2x-x^{2}}+1}=-2. \end{equation*}$$
• For 3) expand the fraction into two and multiply each one by the conjugate of the numerator: $$\begin{eqnarray*} &&\frac{\sqrt{x+2}+\sqrt{x+6}-\sqrt{6}-\sqrt{2}}{x} \\ &=&\sqrt{2}\frac{\left( \sqrt{x/2+1}-1\right) }{x}+\sqrt{6}\frac{\left( \sqrt{x/6+1}-1\right) }{x} \\ &=&\sqrt{2}\frac{\left( \sqrt{x/2+1}-1\right) \left( \sqrt{x/2+1}+1\right) }{x\left( \sqrt{x/2+1}+1\right) }+\sqrt{6}\frac{\left( \sqrt{x/6+1}-1\right)\left( \sqrt{x/6+1}+1\right) }{x\left( \sqrt{x/6+1}+1\right) } \\ &=&\sqrt{2}\frac{1/2}{\sqrt{x/2+1}+1}+\sqrt{6}\frac{1/6}{\sqrt{x/6+1}+1}. \end{eqnarray*}$$ Thus $$\begin{eqnarray*} &&\lim_{x\rightarrow 0}\frac{\sqrt{x+2}+\sqrt{x+6}-\sqrt{6}-\sqrt{2}}{x} \\ &=&\lim_{x\rightarrow 0}\left( \sqrt{2}\frac{1/2}{\sqrt{x/2+1}+1}+\sqrt{6}\frac{1/6}{\sqrt{x/6+1}+1}\right) \\ &=&\frac{1}{12}\sqrt{6}+\frac{1}{4}\sqrt{2}. \end{eqnarray*}$$
1) Factorise both the numerator and denominator so that the terms $(1-x)^n$ can be simplified 2) Use $$\alpha-\beta=\frac{\alpha^2-\beta^2}{\alpha+\beta}$$ 3) Write
$$\sqrt{x + 2} + \sqrt{x + 6} - \sqrt{6} - \sqrt{2}=(\sqrt{x + 2} - \sqrt{2})+ (\sqrt{x + 6} - \sqrt{6})$$ and use what we did in $2$.
4) Use $$\alpha-\beta=\frac{\alpha^3-\beta^3}{\alpha^2+\alpha\beta+\beta^2}$$