Answer to Question #104105 in Calculus for Akshay Kumar

Question

For which values of a and b is the following equation true?

lim (sinx/x³ + a + b/x²) = 0
x - 0

Answer 1

The limit equation is given by:

$\lim_{x \to 0} ( \frac{\sin (x)}{x^3} + a + \frac{b}{x^2} ) = 0$

Separating the constant term $a$ from the limit and taking limit for the other limit function, we have:

$\lim_{x \to 0 } ( \frac{\sin (x)}{x^3} + \frac{b}{x^2}) + a = 0$

Now:

The limit value of the limit function is:

$\lim_{x \to 0} ( \frac{ \sin(x) + bx}{x^3} )$

Inserting $x = 0$ into the limit function, we can see that we have $\frac{0}{0}$ form occurs. For this $\frac{0}{0}$ form, we will use L'Hospital's Rule. Using this rule, we will differentiate both numerator and denominator of the limit function. After differentiating both numerator and denominator of the limit function, we have:

$\lim_{x \to 0} \left( \frac{ \cos (x) + b}{3x^2} \right)$

$= \frac{\cos(0) + b}{3(0^2)} \hspace {1 cm}$ [ By substituting $x = 0$ into the limit function ]

$= \frac{1 + b}{0} \hspace{1 cm}$ [ $\because \cos (0) = 1$ ]

$= \infty$

As, the right hand side of the limit equation is so, we must keep this limit value to . For this, $b + 1$ must be equal to

So:

$b + 1 = 0 \\ \Rightarrow b = -1$

Now:

Inserting $b = -1$ into the limit equation, we have:

$\lim_{x \to 0} ( \frac{\sin (x)}{x^3} - \frac{1}{x^2} ) + a = 0$

$\Rightarrow \lim_{x \to 0} ( \frac{\sin (x) - x}{x^3} ) + a = 0$

$\Rightarrow \lim_{x \to 0} ( \frac{ \cos (x) - 1 }{3x^2} ) + a = 0$ [ $\because \frac{0}{0}$ form occurs so, we use L'Hospital's Rule ]

$\Rightarrow \lim_{x \to 0} ( \frac{ - \sin (x)}{6x} ) + a = 0$ [ Again, $\frac{0}{0}$ form occurs so, we use L'Hospital's Rule ]

$\Rightarrow - \lim_{x \to 0} ( \frac{\cos (x)}{6} ) + a = 0$ [ Again, $\frac{0}{0}$ form occurs so, we use L'Hospital's Rule]

$\Rightarrow - \frac{\cos(0)}{6} + a= 0$ [ Substituting $x = 0$ into the limit function ]

$\Rightarrow - \frac{1}{6} + a = 0$ [ $\because \cos (0) = 1$ ]

$\Rightarrow a = \frac{1}{6}$

Hence:

The values of $a \,\, and \,\, b$ are $\frac{1}{6} \, and \, -1$ respectively.