Ignoring higher order terms in approximations Suppose we're approximating a weird function $f(x)$ using taylor polynomial $p(x)$.
$(x-a)^{100}$ is so tiny when $|x-a| < 1$ so we can ignore the higher order terms.
But what happens to values that are beyond $1$ unit ? Isn't $(x-a)^{100}$ too large and wouldn't the higher order terms suddenly contribute way more than the lower order terms ? I know most of the time we stay around the pivot value. I feel my confusion is in seeing why/how the higher order terms suddenly become dominating as soon as I go a tiny bit beyond $1$ unit ?
 A: To do an approximation you write the taylor polynomial around the point you want to make an approximation. Say you want an approximation at $1.9$ then you will approximate around $2$. We do this because this requires fewer terms in the approximation.
A taylor polynomial is correct upto a certain range, if you keep adding more terms you can use to approximate any value. The higher order terms dominate because you tuncate the polynomial. Consider the following taylor series for $\sin x$ about $0$.
$$\sin(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}\cdots$$
Here I have plotted what the polynomial looks if you take first two terms(blue), first 3 terms(orange), first four terms(green) and sine function(red).

Each higher order term gets balanced by the next terms. As you can see when we add more terms we make the approximation more accurate and also make the range in which approximation works larger. Your $x^{100}$ term is being balanced by $x^{101}$ and by all terms with power less than 100, i.e $x^{99}, x^{98}\cdots$. If you truncate $x^{100}$ the balancing power of the lower term will reduce as the x goes away from the origin. Also consider the coefficient of the terms get smaller and smaller. $x^n$ has $1/n!$ as its coefficient. So $x^{100}$ will become a problem much much later than at $x=1$.
A: Let's look at the Taylor Expansion of $e^x$.
We have $\displaystyle e^x=1+x+\frac{x^2}{2}+\frac{x^3}{6}+\frac{x^4}{24}...$
So no, the upper exponents don't immediately dominate. 
The coefficient of $x^{100}$ would be like $\displaystyle \frac{1}{100!}$ which is a little bit gigantic.
But $e^x$ is supposed to grow big. I'll show you $e^{-x}$ which stays pretty small for positive $x$.
We have $\displaystyle e^{-x}=1-x+\frac{x^2}{2}-\frac{x^3}{6}+\frac{x^4}{24}$.
The positives and negatives almost cancel each other out, despite being very large terms themselves. You are left with a small nonzero number. If $x$ is negative, then the minus terms become positive, so nothing cancels out and you are left with a large number, $e^{|x|}$. 
