**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 \(\infty\).

*Related*

If x tends to infinity then p(x) is actually e raises to x. Then e raise x/p(x)=1 when limit x tends to infinity.

Are you sure that you are taking limit as x tends to infinity? I think you might be taking limit as n tends to infinity.