Computing: limx→0log⁡(1+x)x2−1xCould the following limit be computed without L'Hopital and Taylor? Thanks.

Here's an approach. Note that you can write the limit aslimx→0log⁡(1+x)−xx2and use the following definition:log⁡(1+x)=∫11+xdttFor x small, the midpoint rule gives us that:log⁡(1+x)≈x2(1+11+x)=x+12x21+xand hence we havelimx→0x+12x21+x−xx2=limx→0x+12x2−x−x2x2=limx→0−12x2x2=−t12You may consider the midpoint rule to be a cheat, since it is typically justified using Taylor's theorem (I suspect that it can be proved without Taylor's theorem, though I haven't tried it).

Explained: "Computing: limx→0log⁡(1+x)x2−1xCould the following limit be computed without..."

navantegipowh

Answered question

2022-03-25

Computing:

lim_{x \to 0} \frac{\log (1 + x)}{x^{2}} - \frac{1}{x}

Could the following limit be computed without L'Hopital and Taylor? Thanks.

Answer & Explanation

Yaritza Phillips

Beginner2022-03-26Added 12 answers

Here's an approach. Note that you can write the limit as

lim_{x \to 0} \frac{\log (1 + x) - x}{x^{2}}

and use the following definition:

\log (1 + x) = \int_{1}^{1 + x} \frac{d t}{t}

For x small, the midpoint rule gives us that:

\log (1 + x) \approx \frac{x}{2} (1 + \frac{1}{1 + x}) = \frac{x + \frac{1}{2} x^{2}}{1 + x}

and hence we have

lim_{x \to 0} \frac{\frac{x + \frac{1}{2} x^{2}}{1 + x} - x}{x^{2}} = lim_{x \to 0} \frac{x + \frac{1}{2} x^{2} - x - x^{2}}{x^{2}} = lim_{x \to 0} \frac{- \frac{1}{2} x^{2}}{x^{2}} = - t \frac{1}{2}

You may consider the midpoint rule to be a cheat, since it is typically justified using Taylor's theorem (I suspect that it can be proved without Taylor's theorem, though I haven't tried it).

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