Q.Evaluate: $\\displaystyle\\lim_{x\\to\\infty}\\dfrac{2x^2-3}{x^2-5x+3}$

When computing $\lim_{x\to\alpha}R(x)$, direct substitution can produce one of seven indeterminate forms — expressions that look numeric but cannot be assigned a value by the ordinary rules of arithmetic:

None of these tell you the actual limit — the true value depends on how fast each part approaches its own limit, which is exactly what l'Hôpital's Rule (discovered by Johann Bernoulli, published by Guillaume de l'Hôpital) resolves using derivatives.

l'Hôpital's Rule. Suppose $f(x)$ and $g(x)$ are differentiable with $g'(x)\ne0$ near $x=a$.

• If $\displaystyle\lim_{x\to a}f(x)=0=\lim_{x\to a}g(x)$ (a $\tfrac00$ form), or
• if $\displaystyle\lim_{x\to a}f(x)=\pm\infty=\lim_{x\to a}g(x)$ (a $\tfrac{\infty}{\infty}$ form),

then

(provided the right-hand limit exists), and the rule may be reapplied if the new ratio is again $\tfrac00$ or $\tfrac{\infty}{\infty}$. The rule also applies with $x\to a^{\pm}$ or $x\to\pm\infty$.

Reducing the other five forms to $\tfrac00$ or $\tfrac{\infty}{\infty}$ first:

• $0\times\infty$: rewrite the product $f\cdot g$ (with $f\to0,\,g\to\infty$) as $\dfrac{f}{1/g}$ (a $\tfrac00$ form) or $\dfrac{g}{1/f}$ (a $\tfrac{\infty}{\infty}$ form).
• $\infty-\infty$: combine the two terms into a single fraction (common denominator); the combined expression is then usually $\tfrac00$ (or simplifies algebraically before any limit rule is needed).
• $0^0,\,1^{\infty},\,\infty^0$ (all come from an expression $g(x)^{h(x)}$): let $y=g(x)^{h(x)}$, take $\log$ to get $\log y=h(x)\log g(x)$ — a $0\times\infty$ form — reduce that to $\tfrac00$/$\tfrac{\infty}{\infty}$ and apply l'Hôpital, then exponentiate the resulting value of $\lim\log y$ back (using $\lim\log y=\log\left(\lim y\right)$ for continuous $\log$) to get $\lim y=e^{\text{(that value)}}$.