Find the limit of ln(100x^100 + 5x + 3) / ln(2x^20 + 3x^2 - 2) as x tends to infinity.

Step-by-step solution

1. 1

   Identify the form. Both logarithms tend to $+\infty$, so the quotient is of type $\tfrac{\infty}{\infty}$. L'Hopital would work but produces a mess of rational functions; exploiting the structure of $\ln$ is far cleaner.

2. 2

   Replace each polynomial by its leading term. As $x\to\infty$,

   $100x^{100}+5x+3\sim100x^{100},\qquad2x^{20}+3x^{2}-2\sim2x^{20}$

   This substitution is legitimate inside a logarithm because the neglected part contributes $\ln\left(1+o(1)\right)\to0$, an additive term that vanishes — not a multiplicative error.

3. 3

   Split each logarithm with $\ln(ab)=\ln a+\ln b$.

   $\ln\left(100x^{100}\right)=\ln100+100\ln x,\qquad\ln\left(2x^{20}\right)=\ln2+20\ln x$

   The exponents come down as multipliers of $\ln x$ — this is where the degrees $100$ and $20$ take over.

4. 4

   Divide numerator and denominator by $\ln x$. Since $\ln x\to\infty$:

   $\lim_{x\to\infty}\frac{\frac{\ln100}{\ln x}+100}{\frac{\ln2}{\ln x}+20}=\frac{0+100}{0+20}=5$

   The leading coefficients $100$ and $2$ disappear entirely — inside a logarithm a constant factor is only an additive constant, negligible beside $\ln x$.

5. 5

   State the general rule and verify. For polynomials of degrees $p$ and $q$, the limit of the log ratio is simply $\tfrac{p}{q}$ — here $\tfrac{100}{20}=5$ ✓. Numerically the expression gives $5.0244$ at $x=10$, $5.0123$ at $x=100$, $5.0025$ at $x=10^{10}$ and $5.00025$ at $x=10^{100}$ ✓ — slow but unmistakable convergence to $5$, the slowness being characteristic of logarithmic limits.