Question

Lim x->0 (√x+9-3/sin6x)​

Answer 1

Ответ:

Пошаговое объяснение:

здесь двустороннего предела нет

предел справа

$\displaystyle \lim_{x \to 0^{+}} \bigg (\sqrt{x} +9-\frac{3}{sin6x} \bigg ) = 0+9 -\lim_{x \to 0^{+}}\frac{1}{2x} = 9-\infty= - \infty$

здесь предел 2х при х →0 равен 0 и 2х > 0 lдля всех х >0 поэтому

$\displaystyle \lim_{x\to 0^{+}} \frac{1}{2x} = \infty$

предел слева

$\displaystyle \lim_{x \to 0^{-}} \bigg (\sqrt{x} +9-\frac{3}{sin6x} \bigg ) = 0+9 -\lim_{x \to 0^{-}}\frac{1}{2x} = 9-(-\infty)= \infty$

здесь предел 2х при х →0 равен 0 и 2х < 0 lдля всех х <0 поэтому

$\displaystyle \lim_{x\to 0^{-}} \frac{1}{2x} =- \infty$

Answer 2

To find the limit of the expression as x approaches 0:

lim (x->0) (√(x+9) - 3/sin(6x))

First, we can simplify the expression by factoring out the common factor of 3 in the numerator:

lim (x->0) (3(√(x+9)/3) - 3/sin(6x))

Now, we can use the limit properties:

lim (x->0) [3(√(x+9)/3)] - lim (x->0) (3/sin(6x))

Notice that as x approaches 0, the first term inside the square brackets simplifies to:

(3/3)lim (x->0) √(x+9) = lim (x->0) √(x+9)

Now, let's consider the second term:

lim (x->0) (3/sin(6x))

This limit requires some additional work. We can use the fact that sin(x) is approximately equal to x for small x (this is known as the small-angle approximation):

lim (x->0) (3/sin(6x)) ≈ lim (x->0) (3/(6x))

Now, we can simplify further:

lim (x->0) (3/(6x)) = (1/2)lim (x->0) (1/x)

As x approaches 0, the limit of 1/x is infinite (∞). So:

(1/2)lim (x->0) (1/x) = (1/2)∞ = ∞

Now, we have simplified both terms, and the expression becomes:

lim (x->0) √(x+9) - ∞

As x approaches 0, √(x+9) approaches √9 = 3. Therefore:

lim (x->0) √(x+9) - ∞ = 3 - ∞

Since we have infinity as a result, this limit is undefined. In other words, as x approaches 0, the expression approaches negative infinity (∞), and it does not have a finite limit.