Answer to Question #105777 in Calculus for khushi

Question

For which values of a and b is the following equation true?
lim((sin(2x)/x^3)+a+(b/x^2))=0,x approaches to 0

Answer 1

The limit equation is given by:

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

Separating the constant $a$ from the limit, we have:

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

$\Rightarrow \lim_{x \to 0} \left( \frac{ \sin (2x) + bx }{x^3} \right) + a = 0$

After substituting the limit value $x = 0$ into the limit function, we can see that we have $\frac{0}{0}$ form is exist. So we will apply L'Hospital's Rule to this limit function.

Now:

We know that this rule states that if such indeterminate form $\frac{0}{0}$ exists then we will take the derivatives of the numerator and denominator and substitute the limit value. This process will apply again and again until we get a finite value.

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Applying L'Hospital's Rule to the limit function, we have:

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

Substituting $x = 0$ into the limit function, we have:

$\Rightarrow \left( \frac{2 + b}{3(0^2)} \right) + a \hspace{1 cm} \left[ \because \cos (0) = 1 \right]$

After substituting $x = 0$ into the limit function, we get $\infty$ . But this result is not possible because the right hand side of the limit equation is zero. So, we have to find a value of $b$ so that $b + 2 = 0 \, or \, b = - 2$

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Now:

Substituting $b = - 2$ into the limit function we have:

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

$\Rightarrow \frac{2}{3} \lim_{x \to 0} \left( \frac{\cos (2x) - 1}{x^2} \right) + a = 0$

$\Rightarrow \frac{2}{3} \lim_{x \to 0} \left( \frac{ - 2 \sin (2x) - 0}{2x} \right) + a = 0$ $\hspace{1 cm}$ [ $\because \frac{0}{0}$ form, so we used L'Hospital's Rule ]

$\Rightarrow - \frac{2}{3} \lim_{x \to 0} \left( \frac{ \sin (2x)}{x} \right) + a = 0$

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

$\Rightarrow - \frac{2}{3} \left( 2 \cos(0) \right) + a = 0 \hspace{1 cm}$ [ Substitute $x = 0$ into the limit function ]

$\Rightarrow - \frac{4}{3} + a = 0 \hspace{1 cm} \left[ \because \cos(0) = 1 \right]$

$\Rightarrow a = \frac{4}{3}$

Hence:

The values of $a\, \, and \,\, b$ are $\frac{4}{3} \, and \, -2$ respectively.