Question

Lim x→+∞ ((0,5x^2+x)*exp(-3*x))

Answer 1

Ответ:

Объяснение:

(0.5x²+x)/e³ˣ

lim x→+∞

имеем неопределенность вида ∞/∞

правило Лопиталя для такого случая:

предел отношения равен пределу отношения производных от числ и знаменателя.

(x+1)/3e³ˣ

lim x→+∞

то же самое....  ∞/∞

второй раз применим

1/9e³ˣ        = 1/∞ = 0

lim x→+∞

Answer 2

To find the limit of the function $f(x) = (0.5x^2 + x)e^{-3x}$ as $x$ approaches positive infinity, we'll analyze the behavior of the function as $x$ gets very large.

First, let's look at the two terms in the expression:

1. $0.5x^2$ grows much faster than $x$ as $x$ becomes very large, because the exponent is 2. So, for large $x$, $0.5x^2$ will dominate the behavior of the function.

2. $e^{-3x}$ approaches 0 as $x$ approaches positive infinity. This is because the exponential function $e^{-3x}$ decreases very rapidly as $x$ increases.

Since $e^{-3x}$ approaches 0 and $0.5x^2$ dominates the behavior, we have:

$\lim_{x\to+\infty} (0.5x^2 + x)e^{-3x} = \lim_{x\to+\infty} 0.5x^2 \cdot e^{-3x}$

Now, we can apply the limit properties:

$\lim_{x\to+\infty} 0.5x^2 \cdot e^{-3x} = 0.5 \lim_{x\to+\infty} x^2 \cdot \lim_{x\to+\infty} e^{-3x} = 0$

Since both $x^2$ and $e^{-3x}$ approach 0 as $x$ goes to positive infinity, the overall limit of the function $f(x)$ as $x$ approaches positive infinity is 0.