Evaluate the following limits :
`lim_(theta to pi/2)(tan2theta)/(theta-pi/2)`

Answer

To evaluate the limit

\[ \lim_{\theta \to \frac{\pi}{2}} \frac{\tan(2\theta)}{\theta - \frac{\pi}{2}}, \]

we will follow these steps:

### Step 1: Rewrite the limit
We notice that as \(\theta\) approaches \(\frac{\pi}{2}\), the denominator approaches \(0\). We can rewrite the limit in a more manageable form. We can express \(\theta\) in terms of a new variable \(h\) where \(h = \theta - \frac{\pi}{2}\). Therefore, as \(\theta\) approaches \(\frac{\pi}{2}\), \(h\) approaches \(0\).

### Step 2: Substitute for \(\theta\)
Now, we can express \(\theta\) as:

\[ \theta = h + \frac{\pi}{2}. \]

Substituting this into our limit gives:

\[ \lim_{h \to 0} \frac{\tan(2(h + \frac{\pi}{2}))}{h}. \]

### Step 3: Simplify the argument of the tangent
Using the angle addition formula for tangent, we have:

\[ \tan(2(h + \frac{\pi}{2})) = \tan(2h + \pi) = \tan(2h). \]

This is because \(\tan(x + \pi) = \tan(x)\).

### Step 4: Substitute back into the limit
Now our limit becomes:

\[ \lim_{h \to 0} \frac{\tan(2h)}{h}. \]

### Step 5: Apply the limit property
We can use the limit property that states:

\[ \lim_{x \to 0} \frac{\tan(kx)}{kx} = 1. \]

In our case, we can set \(k = 2\):

\[ \lim_{h \to 0} \frac{\tan(2h)}{2h} \cdot 2 = 1 \cdot 2 = 2. \]

### Final Result
Thus, we conclude that:

\[ \lim_{\theta \to \frac{\pi}{2}} \frac{\tan(2\theta)}{\theta - \frac{\pi}{2}} = 2. \]