Please help. I do not understand this solution. Linear independence. Show that $\{f_a : (0,1)\rightarrow R | f_a(x)=\dfrac{1}{1-ax}\}_{a\in(0,1)}$ is linearly independent. 
Here is the solution from the textbook: 
Let n is a positive integer. Suppose $a_1,\dots,a_n \in (0,1)$ and $ a_i\ne a_j (i\ne j)$ that $\sum\limits_{i=1}^{n}\lambda_i f_i=0, \quad \forall x\in(0,1)$
$\forall x \in (0,1): \sum\limits_{i=1}^{n}\dfrac{\lambda_i}{1-a_ix}=0$
=>$\forall x \in (0,1): \lambda_i=-(1-a_ix)\sum\limits_{1\geq j \leq n (i\ne j)}\dfrac{\lambda_j}{1-a_jx}$.
Let x approach $\dfrac{1}{a_i}$ then $\lambda_i=0$
My question is: since $x\in (0,1)$ so does it make sense to make x tend to $\dfrac{1}{a_i}>1$ ? If it does not work, please show me how to solve this problem. Thank you !
 A: Let n is a positive integer. Suppose $a_1,\dots,a_n \in (0,1)$ and $ a_i\ne a_j (i\ne j)$ that $\sum\limits_{i=1}^{n}\lambda_i f_i=0, \quad \forall x\in(0,1)$
Now, choose  $x_1,\dots,x_n \in (0,1)$ and $ x_i = a_i $. The equation $\sum\limits_{i=1}^{n}\lambda_i f_i=0, \; \forall x_j $ will give you the folowing matrix equation,
$$\mathbf{A}\vec{\lambda} = 0,\quad A_{ij} = \frac{1}{1-a_ia_j}$$
Since 
$$\det A= \frac{\prod_{{\substack{
   i,j \\
   i<j
  }}}(a_i-a_j)^2}{\prod_{{\substack{
   i,j \\
   i\leq j
  }}}(a_ia_j-1)^2} \neq 0$$
You can conclude that $\vec{\lambda} =0, \forall n$. 
(Note, to find the determinant, you can use induction as in the proof of cauchy determinant.
If you accept the formula of Cauchy Determinant, the matrix $A_{ij} = \frac{\frac{1}{a_i}}{\frac{1}{a_i}-a_j}$ is very close to the Cauchy Matrix. When calculating $\det A$, you can factor out $\frac{1}{a_i}$ from the numerator in each column and what's left is exactly a Cauchy Determinant.)
A: Make induction on $n$; the base case is clear because those functions are nonzero.
So assume
$$
\sum_{i=1}^{n+1}\lambda_if_{a_i}(x)=0
$$
If $\lambda_{n+1}=0$, we're done; so it's not restrictive to assume $\lambda_{n+1}=-1$, so that
$$
\frac{1}{1-a_{n+1}x}=\sum_{i=1}^n\frac{\lambda_i}{1-a_ix}
$$
Multiply both sides by $1-a_{n+1}x$, so
$$
1=\sum_{i=1}^n\frac{\lambda_i(1-a_{n+1}x)}{1-a_ix}
$$
This is an equality between rational functions that is satisfied for all $x\in(0,1)$. Let $P(x)=(1-a_1x)(1-a_2x)\dots(1-a_nx)$, so we have
$$
P(x)=\sum_{i=1}^n \lambda_i(1-a_{n+1}x)P_i(x)
$$
where $P_i(x)=P(x)/(1-a_i(x))$ is a polynomial.
Since this equality is satisfied by infinitely many values of $x$ the two polynomials are equal. But the roots of $P$ are $1/a_1,1/a_2,\dots,1/a_n$, while the right hand side has $1/a_{n+1}$ as root. This is a contradiction.

The method used in your textbook would require some more words: the function you're using is the restriction of a continuous function defined on a larger set (where $x\ne a_i^{-1}$), so it's legitimate to draw conclusions on the restriction from the behavior of the “more general” function.
The algebraic method I used is based on the same idea, it just doesn't require limits.
