Submitted by olliecook1004 on Tue, 09/12/2017 - 18:32

Forums:

The first part was fine, then it asked 'Deduce that if x is very large....' I have no idea where to start, can anyone explain this?

Submitted by olliecook1004 on Tue, 09/12/2017 - 18:32

Forums:

The first part was fine, then it asked 'Deduce that if x is very large....' I have no idea where to start, can anyone explain this?

## If $x$ is very large, there

If $x$ is very large, there are some approximations we can take. For example $$2x+\frac{1}{x}\approx 2x,\quad x^5+50x^3+1000\approx x^5.$$ In general these approximations use the fact that, for large $x$, an expression that is a sum of powers of $x$ (some possibly negative powers) are dominated by the term with the highest power of $x$ i.e. this term is much bigger than any of the other terms. Can you see how you can apply this idea to the $\sqrt{1+x^2}$ in the denominator?

## That makes sense!

Perfect, the route of 1+x^2 is approximately equal to x. So the route of 1+x^2 minus x is a very small number. 1/2x is a very small number. Correct?

## Sort of

You cannot really say that two very small numbers must be approximately equal. 0.000001 and 0.000000001 are both very small but one is 1000 times as big as the other.

We have $\sqrt{1+x^2} - x = \dfrac 1 {\sqrt{1+x^2} + x}$. The RHS of this is:

\[

\dfrac 1 {\sqrt{1+x^2} + x} \approx \dfrac 1 {x+x}

\]

Hence $\sqrt{1+x^2} - x \approx \dfrac 1 {2x}$.