Math Problem Statement

Вычислить предел двумя способами: используя правило Лопиталя и формулы Тейлора. Если предел не при x → 0 , то для применения формул Тейлора не забудьте предварительно выполнить замену переменной, чтобы получить предел при x → 0

Вычислить предел двумя способами: используя правило Лопиталя и формулы Тейлора. Если предел не при x → 0, то для применения формул Тейлора не забудьте предварительно выполнить замену переменной, чтобы получить предел при x → 0.

Solution

To solve this limit using L'Hôpital's rule and Taylor series expansion, let us analyze the problem and proceed step by step.

Problem:

We are tasked with evaluating the following limit: limx3cosxln(x3)ln(exe3)\lim_{x \to 3} \frac{\cos x \ln(x - 3)}{\ln(e^x - e^3)}

We will:

  1. Use L'Hôpital's Rule.
  2. Use Taylor series expansions, possibly after a substitution to simplify the expressions.

1. Using L'Hôpital's Rule:

First, analyze the behavior of the numerator and denominator as x3x \to 3:

  • Numerator: cosxln(x3)cos3ln0cos3()\cos x \ln(x - 3) \to \cos 3 \cdot \ln 0 \to \cos 3 \cdot (-\infty). However, ln(x3)\ln(x - 3) \to -\infty dominates here.
  • Denominator: ln(exe3)ln0\ln(e^x - e^3) \to \ln 0 \to -\infty.

So, the limit is of the indeterminate form \frac{-\infty}{-\infty}. We can apply L'Hôpital's Rule.

Step 1: Differentiate numerator and denominator.

Let: f(x)=cosxln(x3),g(x)=ln(exe3).f(x) = \cos x \ln(x - 3), \quad g(x) = \ln(e^x - e^3).

Numerator: Using the product rule, f(x)=sinxln(x3)+cosxx3.f'(x) = -\sin x \ln(x - 3) + \frac{\cos x}{x - 3}.

Denominator: Using the chain rule, g(x)=ddxln(exe3)=exexe3.g'(x) = \frac{d}{dx} \ln(e^x - e^3) = \frac{e^x}{e^x - e^3}.

Step 2: Rewrite the limit:

limx3f(x)g(x)=limx3sinxln(x3)+cosxx3exexe3.\lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{-\sin x \ln(x - 3) + \frac{\cos x}{x - 3}}{\frac{e^x}{e^x - e^3}}.

Substitute x=3x = 3 into the new expression to check if further differentiation is needed.

  1. Evaluate each term to simplify further.

2. Using Taylor Series Expansion:

Using Taylor series, we need to simplify the expressions ln(x3)\ln(x - 3) and ln(exe3)\ln(e^x - e^3). For this purpose, we can make the substitution t=x3t = x - 3, which transforms the limit to t0t \to 0.

Step 1: Substitution t=x3t = x - 3, so x=t+3x = t + 3:

Rewriting the terms:

  • ln(x3)=lnt\ln(x - 3) = \ln t,
  • cosx=cos(t+3)cos3sin3t+cos3t22\cos x = \cos(t + 3) \approx \cos 3 - \sin 3 \cdot t + \frac{-\cos 3 \cdot t^2}{2} (Taylor expansion around t=0t = 0),
  • ln(exe3)=ln(et+3e3)=ln(e3(et1))=lne3+ln(et1)\ln(e^x - e^3) = \ln(e^{t + 3} - e^3) = \ln(e^3(e^t - 1)) = \ln e^3 + \ln(e^t - 1).

Simplify ln(et1)\ln(e^t - 1) using its Taylor expansion around t=0t = 0: ln(et1)ln(t+t22+)lnt.\ln(e^t - 1) \approx \ln(t + \frac{t^2}{2} + \dots) \approx \ln t.

Step 2: Rewrite the limit:

After substitution and simplifications, the Taylor expansions help resolve the behavior of the numerator and denominator.


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Math Problem Analysis

Mathematical Concepts

Limits
L'Hôpital's Rule
Taylor Series
Substitution in Limits

Formulas

L'Hôpital's Rule: \( \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \), if \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)
Taylor Series Expansion: \( f(x) \approx f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \dots \)

Theorems

L'Hôpital's Rule
Taylor's Theorem

Suitable Grade Level

Undergraduate