Question 18 of 70intermediate🔧 ApplyNumerical1 mark

Evaluate the following limits in Exercises 1 to 22. lim⁡x→0sin⁡ax+bxax+sin⁡bxa,b,a+b≠0\lim_{x \to 0} \frac{\sin ax + bx}{ax + \sin bx} a, b, a + b \neq 0

Correct Answer

1

Exercise: EXERCISE 12.1 | Q: 20 | (Page Page 22)
For More Understanding

Explanation

To solve this limit, we first check the value of the expression at x = 0. Substituting x = 0 yields the indeterminate form 0/0. To resolve this, we divide both the numerator and the denominator by x. This allows us to separate the terms and utilize the standard limit theorem provided in the context: lim (x→0) (sin x)/x = 1. By adjusting the arguments of the sine functions to match the denominators (multiplying by a/b factors where necessary), we evaluate the limits of individual terms to arrive at the final result.

Solution Steps

  1. Step 1: Check the form of the limit by substituting x=0x = 0. Numerator: sin⁡(a(0))+b(0)=0\sin(a(0)) + b(0) = 0. Denominator: a(0)+sin⁡(b(0))=0a(0) + \sin(b(0)) = 0. Since the form is 00\frac{0}{0}, we proceed by dividing the numerator and denominator by xx.

  2. Step 2: Divide the numerator and denominator by xx.

    lim⁡x→0sin⁡axx+bxxaxx+sin⁡bxx=lim⁡x→0sin⁡axx+ba+sin⁡bxx\lim_{x \to 0} \frac{\frac{\sin ax}{x} + \frac{bx}{x}}{\frac{ax}{x} + \frac{\sin bx}{x}} = \lim_{x \to 0} \frac{\frac{\sin ax}{x} + b}{a + \frac{\sin bx}{x}}

  3. Step 3: Rearrange the terms to apply the standard limit lim⁡x→0sin⁡xx=1\lim_{x \to 0} \frac{\sin x}{x} = 1. Multiply and divide the sine terms by their angles:

    =lim⁡x→0a(sin⁡axax)+ba+b(sin⁡bxbx)= \lim_{x \to 0} \frac{a \left( \frac{\sin ax}{ax} \right) + b}{a + b \left( \frac{\sin bx}{bx} \right)}

  4. Step 4: Evaluate the limit using Theorem 5. As x→0x \to 0, ax→0ax \to 0 and bx→0bx \to 0. Therefore, sin⁡axax→1\frac{\sin ax}{ax} \to 1 and sin⁡bxbx→1\frac{\sin bx}{bx} \to 1.

    =a(1)+ba+b(1)= \frac{a(1) + b}{a + b(1)}

  5. Step 5: Simplify the expression.

    =a+ba+b=1= \frac{a + b}{a + b} = 1

    (Since a+b≠0a + b \neq 0, division is defined).

PreviousNext