Select Page

Question:

True/False?

Let $$P(x)= 1+x+ \frac{x^2}{2!} +… \frac{x^n}{n!}$$ where n is a large positive integer. Then $$\lim_{x\to\infty} \frac{e^x}{P(x)} = 1$$

Hint:

n really doesn’t matter!

Discussion:

As $$x$$ approaches ‘infinity’, both $$e^x$$ and $$P(x)$$ tends to ‘infinity’ (i.e, gets arbitrarily large).

One could use L’Hospitals rule here. Without computation, the numerator $$e^x$$ when differentiated gives $$e^x$$ again. The denominator $$P(x)$$ when differentiated will give a polynomial of degree $$n-1$$. If $$n-1=0$$, i.e, the polynomial obtained by differentiating is a constant one, then the limit is simply limit of $$e^x$$ which is $$\infty$$. If $$n-1 \neq 0$$ then again L’Hospital rule is applicable since any non-constant polynomial has limit $$\infty$$.

Repeating the process sufficiently many times we will obtain a stage where numerator as always will remain $$e^x$$ and denominator will be some constant polynomial. So the limit will be $$\infty$$.