Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
3  
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). –  Matt 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
1  
@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

2 Answers 2

up vote 1 down vote accepted
  • 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 $$\begin{eqnarray*} &&\lim_{x\rightarrow 0}\frac{\sqrt{x+2}+\sqrt{x+6}-\sqrt{6}-\sqrt{2}}{x} \\ &=&\lim_{x\rightarrow 0}\left( \sqrt{2}\frac{1/2}{\sqrt{x/2+1}+1}+\sqrt{6}\frac{1/6}{\sqrt{x/6+1}+1}\right) \\ &=&\frac{1}{12}\sqrt{6}+\frac{1}{4}\sqrt{2}. \end{eqnarray*}$$
share|improve this answer

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 $$\alpha-\beta=\frac{\alpha^3-\beta^3}{\alpha^2+\alpha\beta+\beta^2}$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.