Question

Lim х стремиться +бесконечность = 2х3-х+5/3х3+х2+4

Answer 1

Ответ:

\lim_{n \to \infty} \frac{x^2-4x-5}{x^2-2x-3}= [\frac{\infty}{\infty}]=\lim_{n \to \infty} \frac{ \frac{x^2}{x^2}- \frac{4x}{x^2}- \frac{5}{x^2}}{ \frac{x^2}{x^2}- \frac{2x}{x^2}- \frac{3}{x^2}}= \lim_{n \to \infty} \frac{1- \frac{4}{x}- \frac{5}{x^2}}{1- \frac{2}{x}- \frac{3}{x^2}}=\\\\
 \lim_{n \to \infty} \frac{1-0-0}{1-0-0}=1

Второй вариант решения:

\lim_{n \to \infty} \frac{x^2-4x-5}{x^2-2x-3}= [\frac{\infty}{\infty}]= \lim_{n \to \infty} \frac{x^2-4x}{x^2-2x}= \lim_{n \to \infty} \frac{x(x-4)}{x(x-2)} =\\\\ \lim_{n \to \infty}  \frac{x-4}{x-2}= \lim_{n \to \infty} \frac{x}{x}=1  

Третий вариант решения:

\lim_{n \to \infty} \frac{x^2-4x-5}{x^2-2x-3}=[\frac{\infty}{\infty}]= \lim_{n \to \infty} \frac{(x^2-4x-5)'}{(x^2-2x-3)'}=  \lim_{n \to \infty} \frac{(2x-4)'}{(2x-2)'}= \frac{2}{2}=1

Answer 2

To evaluate the limit as x approaches positive infinity of the given expression, we need to examine the behavior of the expression as x becomes larger and larger.

The expression is:

lim (x → +∞) [2x^3 - x + 5] / [3x^3 + x^2 + 4]

To find the limit, we look at the highest power of x in both the numerator and the denominator. In this case, both the numerator and denominator have the highest power of x as x^3.

So, we divide both the numerator and denominator by x^3 to simplify the expression:

lim (x → +∞) [2x^3 - x + 5] / [3x^3 + x^2 + 4] = lim (x → +∞) [2 - (1/x^2) + (5/x^3)] / [3 + (1/x) + (4/x^3)]

As x approaches positive infinity, 1/x^2 and 1/x^3 approach 0. Similarly, 1/x approaches 0 as x becomes larger. Therefore, the expression becomes:

= [2 - 0 + 0] / [3 + 0 + 0] = 2 / 3

So, the limit as x approaches positive infinity of the given expression is 2/3.