Evaluate the following (10-17) limits :
`lim_(y to 0)((x+y)sec(x+y)-xsecx)/y`.

Answer

To evaluate the limit

\[ \lim_{y \to 0} \frac{(x+y) \sec(x+y) - x \sec x}{y}, \]

we can follow these steps:

### Step 1: Check the form of the limit
First, we substitute \(y = 0\):

\[ \frac{(x+0) \sec(x+0) - x \sec x}{0} = \frac{x \sec x - x \sec x}{0} = \frac{0}{0}. \]

Since we have a \(0/0\) indeterminate form, we can apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
According to L'Hôpital's Rule, if we have a limit of the form \(0/0\) or \(\infty/\infty\), we can take the derivative of the numerator and the derivative of the denominator:

\[ \lim_{y \to 0} \frac{(x+y) \sec(x+y) - x \sec x}{y} = \lim_{y \to 0} \frac{d}{dy} \left[(x+y) \sec(x+y) - x \sec x\right] \Big/ \frac{d}{dy}[y]. \]

### Step 3: Differentiate the numerator
We differentiate the numerator using the product rule:

Let \(u = x+y\) and \(v = \sec(x+y)\).

Using the product rule:

\[ \frac{d}{dy}[(x+y) \sec(x+y)] = \sec(x+y) + (x+y) \sec(x+y) \tan(x+y) \cdot \frac{d}{dy}(x+y). \]

Since \(\frac{d}{dy}(x+y) = 1\), we have:

\[ \frac{d}{dy}[(x+y) \sec(x+y)] = \sec(x+y) + (x+y) \sec(x+y) \tan(x+y). \]

Now, we differentiate the second term \(x \sec x\) with respect to \(y\), which is constant with respect to \(y\), so its derivative is \(0\).

Thus, the derivative of the numerator is:

\[ \sec(x+y) + (x+y) \sec(x+y) \tan(x+y). \]

### Step 4: Differentiate the denominator
The derivative of the denominator \(y\) is simply \(1\).

### Step 5: Substitute back into the limit
Now we substitute back into the limit:

\[ \lim_{y \to 0} \left[\sec(x+y) + (x+y) \sec(x+y) \tan(x+y)\right]. \]

### Step 6: Evaluate the limit as \(y \to 0\)
As \(y\) approaches \(0\):

\[ \sec(x+y) \to \sec x, \quad \tan(x+y) \to \tan x, \quad \text{and } (x+y) \to x. \]

Thus, we have:

\[ \sec x + x \sec x \tan x. \]

### Final Result
Therefore, the limit evaluates to:

\[ \sec x + x \sec x \tan x. \]