# TIFR 2013 problem 18 | Problem based on Limit

Try this problem from TIFR 2013 problem 18 based on limit.

Question: TIFR 2013 problem 18

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 $\infty$.

### 2 comments on “TIFR 2013 problem 18 | Problem based on Limit”

1. 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.

1. 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.

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