"Use the method of your choice to evaluate..." was answered by our experts

Tyra

Answered question

2020-11-27

Use the method of your choice to evaluate the following limits.

lim_{(x, y) \to (0, π / 2)} \frac{1 - \cos x y}{4 x^{2} y^{3}}

nitruraviX

Skilled2020-11-28Added 101 answers

We have to solve the limit:

lim_{(x, y) \to (0, π / 2)} \frac{1 - \cos x y}{4 x^{2} y^{3}}

We know the formula,

\cos^{2} x - \sin^{2} x = \cos 2 x

1 - \cos 2 x = 2 \sin^{2} x

for numerator,

1 - \cos x y = 2 \sin^{2} (\frac{x y}{2})

Therefore,

lim_{(x, y) \to (0, \frac{π}{2})} \frac{1 - \cos x y}{4 x^{2} y^{3}} = lim_{(x, y) \to (0, \frac{π}{2})} \frac{2 \sin^{2} (\frac{x y}{2})}{4 x^{2} y^{3}}

We know the formula

lim_{x \to 0} \frac{\sin x}{x} = 1

lim_{x \to 0} \frac{\sin^{2} x}{x^{2}} = 1

Multiplying and dividing by

(\frac{x y}{2})^{2}

lim_{(x, y) \to (0, \frac{π}{2})} \frac{2 \sin^{2} (\frac{x y}{2})}{4 x^{2} y^{3}} = lim_{(x, y) \to (0, \frac{π}{2})} \frac{2 \sin^{2} (\frac{x y}{2})}{(\frac{x y}{2})^{2}} \times \frac{(\frac{x y}{2})^{2}}{4 x^{2} y^{3}}

= lim_{(x, y) \to (0, \frac{π}{2})} 1 \times \frac{(\frac{x y}{2})^{2}}{4 x^{2} y^{3}}

= lim_{(x, y) \to (0, \frac{π}{2})} \frac{\frac{x^{2} y^{2}}{4}}{4 x^{2} y^{3}}

= lim_{(x, y) \to (0, \frac{π}{2})} \frac{x^{2} y^{2}}{4 \times 4 x^{2} y^{3}}

= lim_{(x, y) \to (0, \frac{π}{2})} \frac{y^{2}}{16 y^{3}}

= lim_{(x, y) \to (0, \frac{π}{2})} \frac{1}{16 y}

= \frac{1}{16 \times \frac{π}{2}}

= \frac{2}{16 π}

= \frac{1}{8 π}

Hence, value of limit is

\frac{1}{8 π}

Jeffrey Jordon

Expert2022-04-01Added 2605 answers

Answer is given below (on video)

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?