Question

Use a graphing device to determine whether the limit exists. If the limit exists, estimate its value to two decimal places.

$\underset{x\to 0}{\lim }\frac{x^2}{\cos 5x-\cos 4x}$

Short answer

Yes, the limit exists.

The value of limit of the given function is$-0.22$

Step-by-step solution

Step 1. Given information.

The limit of function here given is,

$\underset{x\to 0}{\lim }\frac{x^2}{\cos 5x-\cos 4x}$

Step 2. Apply L’Hopital’s Rule.

$\underset{x\to 0}{\lim }\left(\frac{x^2}{\cos \left(5x\right)-\cos \left(4x\right)}\right)$

Then,

$\begin{array}{c}=\underset{x\to 0}{\lim }\left(\frac{2x}{-5\sin \left(5x\right)+4\sin \left(4x\right)}\right)\\ =\underset{x\to 0}{\lim }\left(\frac{2}{-25\cos \left(5x\right)+16\cos \left(4x\right)}\right)\end{array}$

Plug in the value$x=0$

$\begin{array}{c}=\frac{2}{-25\cos (5\cdot 0)+16\cos (4\cdot 0)}\\ =-\frac{2}{9}\\ \Rightarrow -\mathrm{0.2222....}\end{array}$

Step 3. Draw the graph.

Now, as a graphical verification use a graphing device.

Therefore, from the graph it is determined that

$\underset{x\to 0}{\lim }\left(\frac{x^2}{\cos \left(5x\right)-\cos \left(4x\right)}\right)=-\frac{2}{9}=-0.222$