Proofs and calculation techniques

How to prove these results

The standard proof relies on a change of variables and a bound. For lim x->∞ x^n/e^x = 0, we use the inequality e^x ≥ x^(n+1)/(n+1)! which holds for x ≥ 0 (a consequence of the convexity of the exponential function), giving 0 ≤ x^n/e^x ≤ (n+1)!/x, and the right-hand side tends to 0.

For lim x->+∞ (ln x)/x^n, we set t = ln(x), so x = e^t. The limit becomes lim t->+∞ t/e^(n*t), which is a further application of the previous theorem.

Practical method for an exercise

1. Identify the dominant term, i.e. the one that grows fastest according to the hierarchy ln(x) << x^n << e^x.
2. Factorise the expression in terms of this dominant term.
3. Apply the comparative growth theorems to what remains, which must tend to 0.

Detailed example

Let us calculate lim x → +∞ (x² - e^x)/(x + e^x).

We factor out e^x from both the numerator and the denominator:

(x² - e^x)/(x + e^x) = (x²/e^x - 1)/(x/e^x + 1)

As x²/e^x → 0 and x/e^x → 0, the limit is (0–1)/(0+1) = –1.

Pitfalls to avoid

• Forgetting to check the behaviour of the dominant term before simplifying.
• Applying the theorem directly to an indeterminate form without first factoring it out.
• Confusing x^n with n^x: these are two very different types of growth, as n^x is exponential in nature.