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Ch. 4 - Applications of the Derivative
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 4, Problema 87

Two methods Evaluate the following limits in two different ways: Use the methods of Chapter 2 and use l’Hôpital’s Rule.
lim_x→0 (e²ˣ + 4eˣ - 5) / (e²ˣ - 1)

Guida verificata passo dopo passo
1
First, let's evaluate the limit using algebraic manipulation. Start by substituting x = 0 directly into the expression to check if it results in an indeterminate form. Substitute x = 0 into the expression (e^(2x) + 4e^x - 5) / (e^(2x) - 1).
Notice that substituting x = 0 gives us (1 + 4*1 - 5) / (1 - 1), which is 0/0, an indeterminate form. This suggests that we can use algebraic techniques or l'Hôpital's Rule to evaluate the limit.
For the algebraic method, consider expanding e^(2x) and e^x using their Taylor series expansions around x = 0: e^(2x) ≈ 1 + 2x + 2x^2/2 and e^x ≈ 1 + x + x^2/2. Substitute these approximations into the original expression.
Simplify the expression using the Taylor series expansions: (1 + 2x + 2x^2/2 + 4(1 + x + x^2/2) - 5) / (1 + 2x + 2x^2/2 - 1). Simplify the numerator and the denominator separately.
Now, let's use l'Hôpital's Rule. Since the limit is in the form 0/0, differentiate the numerator and the denominator with respect to x. The derivative of the numerator e^(2x) + 4e^x - 5 is 2e^(2x) + 4e^x, and the derivative of the denominator e^(2x) - 1 is 2e^(2x). Evaluate the limit of the new expression as x approaches 0.

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