Q. 32

Question

Calculate each of the limits in Exercises 21-48. Some of these limits are made easier by L'Hôpital's rule, and some are not.

$\lim_{x \to \infty} \frac{x e^{- 3 x}}{x^{2} + 3 x + 1}$

Step-by-Step Solution

Verified

Answer

The value of limit is $0$

1Step 1. Given information

Given function $\lim_{x \to \infty} \frac{x e^{- 3 x}}{x^{2} + 3 x + 1}$

2Step 2: Apply L-Hospital rule and solve

Calculating, we get

$\lim_{x \to \infty} \frac{x e^{- 3 x}}{x^{2} + 3 x + 1} = \lim_{x \to \infty} \frac{x}{e^{3 x} (x^{2} + 3 x + 1)} \lim_{x \to \infty} \frac{x e^{- 3 x}}{x^{2} + 3 x + 1} = \lim_{x \to \infty} \frac{x}{e^{3 x} (x^{2} + 3 x + 1)} = \lim_{x \to \infty} \frac{1}{e^{3 x} (2 x + 3) + (x^{2} + 3 x + 1) \cdot e^{3 x} \cdot 3} = \lim_{x \to \infty} \frac{1}{e^{3 x} [2 x + 3 + 3 x^{2} + 9 x + 3]} = \lim_{x \to \infty} \frac{1}{e^{3 x} (3 x^{2} + 11 x + 6)} = 0$

Q. 31

Q. 33

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Q. 31

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Q. 33

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Q. 34

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