Evaluate the following limits :
`Lim_(x to 0 ) (1-cosmx)/(1- cos nx)`

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{1 - \cos(mx)}{1 - \cos(nx)}, \]

we can use the trigonometric identity that relates \(1 - \cos \theta\) to \(\sin^2\) as follows:

\[ 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right). \]

### Step 1: Apply the identity

Using the identity, we can rewrite the limit:

\[ 1 - \cos(mx) = 2 \sin^2\left(\frac{mx}{2}\right) \quad \text{and} \quad 1 - \cos(nx) = 2 \sin^2\left(\frac{nx}{2}\right). \]

Thus, the limit becomes:

\[ \lim_{x \to 0} \frac{2 \sin^2\left(\frac{mx}{2}\right)}{2 \sin^2\left(\frac{nx}{2}\right)} = \lim_{x \to 0} \frac{\sin^2\left(\frac{mx}{2}\right)}{\sin^2\left(\frac{nx}{2}\right)}. \]

### Step 2: Simplify the limit

We can further simplify this limit by rewriting it in terms of \(x\):

\[ \lim_{x \to 0} \frac{\sin^2\left(\frac{mx}{2}\right)}{\sin^2\left(\frac{nx}{2}\right)} = \lim_{x \to 0} \left(\frac{\sin\left(\frac{mx}{2}\right)}{\frac{mx}{2}} \cdot \frac{mx}{2}\right)^2 \cdot \left(\frac{\frac{nx}{2}}{\sin\left(\frac{nx}{2}\right)}\right)^2. \]

### Step 3: Use the standard limit

Using the standard limit \(\lim_{u \to 0} \frac{\sin u}{u} = 1\), we can rewrite the limit:

\[ = \left(\lim_{x \to 0} \frac{\sin\left(\frac{mx}{2}\right)}{\frac{mx}{2}}\right)^2 \cdot \left(\lim_{x \to 0} \frac{n^2x^2/4}{\sin^2\left(\frac{nx}{2}\right)}\right)^{-1}. \]

### Step 4: Evaluate the limits

Now, applying the limits:

1. \(\lim_{x \to 0} \frac{\sin\left(\frac{mx}{2}\right)}{\frac{mx}{2}} = 1\)
2. \(\lim_{x \to 0} \frac{\sin\left(\frac{nx}{2}\right)}{\frac{nx}{2}} = 1\)

Thus, we have:

\[ = \left(1\right)^2 \cdot \frac{m^2}{n^2} = \frac{m^2}{n^2}. \]

### Final Answer

Therefore, the limit evaluates to:

\[ \lim_{x \to 0} \frac{1 - \cos(mx)}{1 - \cos(nx)} = \frac{m^2}{n^2}. \]