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Exercise 7.5 · Q9

Q.Evaluate: displaystylelimxtoinftyleft(1+dfrac1xright)x\\displaystyle\\lim_{x\\to\\infty}\\left(1+\\dfrac1x\\right)^x

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Concept understanding — Indeterminate Forms and L'Hopital's Rule

When computing lim⁡x→αR(x)\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:

00,∞∞,0×∞,∞−∞,1∞,00,∞0.\frac00,\quad \frac{\infty}{\infty},\quad 0\times\infty,\quad \infty-\infty,\quad 1^{\infty},\quad 0^0,\quad \infty^0.

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)f(x) and g(x)g(x) are differentiable with g′(x)≠0g'(x)\ne0 near x=ax=a.

  • If lim⁡x→af(x)=0=lim⁡x→ag(x)\displaystyle\lim_{x\to a}f(x)=0=\lim_{x\to a}g(x) (a 00\tfrac00 form), or
  • if lim⁡x→af(x)=±∞=lim⁡x→ag(x)\displaystyle\lim_{x\to a}f(x)=\pm\infty=\lim_{x\to a}g(x) (a ∞∞\tfrac{\infty}{\infty} form),

then

lim⁡x→af(x)g(x)=lim⁡x→af′(x)g′(x)\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{f'(x)}{g'(x)}

(provided the right-hand limit exists), and the rule may be reapplied if the new ratio is again 00\tfrac00 or ∞∞\tfrac{\infty}{\infty}. The rule also applies with x→a±x\to a^{\pm} or x→±∞x\to\pm\infty.

Reducing the other five forms to 00\tfrac00 or ∞∞\tfrac{\infty}{\infty} first:

  • 0×∞0\times\infty: rewrite the product f⋅gf\cdot g (with f→0, g→∞f\to0,\,g\to\infty) as f1/g\dfrac{f}{1/g} (a 00\tfrac00 form) or g1/f\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 00\tfrac00 (or simplifies algebraically before any limit rule is needed). …

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