Question

Решить lim x->0 (sin7x+sin3x)/x*sinx

Answer 1

$\lim_{x\to0}\frac{sin7x+sin3x}{xsinx}=[sin7x \approx 7x;\ sin3x \approx 3x;\ sinx \approx x]= \\ \\ =\lim_{x\to0}\frac{7x+3x}{x*x}=\lim_{x\to0}\frac{10x}{x^2}=\lim_{x\to0}\frac{10}{x}=\frac{10}{0}=\infty$

Answer 2

To solve the limit of the expression as x approaches 0, we can use the properties of limits and trigonometric identities. Let's break down the expression step by step:

lim(x->0) [(sin(7x) + sin(3x)) / (x * sin(x))]

Step 1: Apply the trigonometric identity sin(a + b) = sin(a)cos(b) + cos(a)sin(b) to the numerator:

= lim(x->0) [(sin(7x)cos(3x) + cos(7x)sin(3x)) / (x * sin(x))]

Step 2: Rearrange the terms in the numerator:

= lim(x->0) [(sin(7x)cos(3x) + sin(3x)cos(7x)) / (x * sin(x))]

Step 3: Apply the trigonometric identity sin(2a) = 2sin(a)cos(a) to the numerator:

= lim(x->0) [2sin(7x)cos(3x) + 2sin(3x)cos(7x)] / (x * sin(x))

Step 4: Apply the limit properties and break down the expression further:

= lim(x->0) 2sin(7x)cos(3x) / (x * sin(x)) + lim(x->0) 2sin(3x)cos(7x) / (x * sin(x))

Step 5: Simplify each term separately:

lim(x->0) [2sin(7x)cos(3x) / (x * sin(x))] = = 2 * lim(x->0) [sin(7x)cos(3x) / (x * sin(x))]

lim(x->0) [2sin(3x)cos(7x) / (x * sin(x))] = = 2 * lim(x->0) [sin(3x)cos(7x) / (x * sin(x))]

Step 6: Apply the limit properties again:

= 2 * lim(x->0) [sin(7x) / x] * lim(x->0) [cos(3x) / sin(x)]

• 2 * lim(x->0) [sin(3x) / x] * lim(x->0) [cos(7x) / sin(x)]

Step 7: Evaluate each limit separately:

lim(x->0) [sin(7x) / x] = 7 (using the limit definition of sin(x) / x)

lim(x->0) [cos(3x) / sin(x)] = cos(0) / sin(0) = 1

lim(x->0) [sin(3x) / x] = 3 (using the limit definition of sin(x) / x)

lim(x->0) [cos(7x) / sin(x)] = cos(0) / sin(0) = 1

Step 8: Substitute the values back into the expression:

= 2 * 7 * 1 + 2 * 3 * 1 = 14 + 6 = 20

Therefore, the limit of (sin(7x) + sin(3x)) / (x * sin(x)) as x approaches 0 is 20.