Is the term true? $\frac{\theta}{\theta - 1 } = \frac{1} {\theta-1} + 1$

Question

Why is the following term true? What is the corresponding rule or how is it transformed?

$\frac{\theta}{\theta - 1 } = \frac{1} {\theta-1} + 1$

Thanks!!

Answer 1

$$\frac{\theta}{\theta - 1 } = \frac{1 + \theta - 1}{\theta - 1 } = \frac{1}{\theta - 1 } + \frac{\theta - 1}{\theta - 1 } = \frac{1} {\theta-1} + 1$$

Answer 2

$$\frac{1}{\theta-1}+1=\frac{1+(\theta-1)\cdot1}{\theta-1}=\frac{1+\theta-1}{\theta-1}=\frac{\theta}{\theta-1}$$

Answer 3

A more-or-less general rule is to observe that $$ \frac{\theta}{\theta-1}\approx 1 $$ so we could try to represent $$ \frac{\theta}{\theta-1}=1+A $$ for some $A$. Doing so, we find that $A=1/(\theta-1)$.

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In a similar vein, if we're given $$ \frac{\theta^2+2}{\theta-1} $$ we could write $$ \frac{\theta^2+2}{\theta-1}=\theta+A $$ Solving for $A$ we find that $$ \frac{\theta^2+2}{\theta-1}=\theta+\frac{\theta+2}{\theta-1} $$ If we wish, we could do a similar construction on the rightmost term to obtain $$ \frac{\theta^2+2}{\theta-1}=\theta+1-\frac{3}{\theta-1} $$ Added. Which, by the way, shows that $y=x+1$ is an asymptote of $y=(x^2+2)/(x-1)$.