Find the asymptotes of the following curves :
`y = x+ (sin x)//x`

Answer

To find the asymptotes of the curve given by the equation \( y = x + \frac{\sin x}{x} \), we will analyze the behavior of the function as \( x \) approaches infinity and zero.

### Step 1: Analyze the function as \( x \to \infty \)

We start by examining the term \( \frac{\sin x}{x} \). As \( x \) approaches infinity, we know that \( \sin x \) oscillates between -1 and 1. Therefore, we can bound the term:

\[ -\frac{1}{x} \leq \frac{\sin x}{x} \leq \frac{1}{x} \]

As \( x \to \infty \), both \( -\frac{1}{x} \) and \( \frac{1}{x} \) approach 0. Thus, by the Squeeze Theorem:

\[ \lim_{x \to \infty} \frac{\sin x}{x} = 0 \]

Now substituting this back into our function:

\[ \lim_{x \to \infty} y = \lim_{x \to \infty} \left( x + \frac{\sin x}{x} \right) = \lim_{x \to \infty} x + 0 = \infty \]

### Step 2: Identify the oblique asymptote

Next, we can find the oblique asymptote by considering the linear part of the function as \( x \to \infty \):

\[ y \approx x \quad \text{as } x \to \infty \]

Thus, the oblique asymptote is the line:

\[ y = x \]

### Step 3: Analyze the function as \( x \to 0 \)

Now, we analyze the function as \( x \) approaches 0. We need to evaluate \( \frac{\sin x}{x} \) at this limit:

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

So substituting this into our function:

\[ \lim_{x \to 0} y = \lim_{x \to 0} \left( x + \frac{\sin x}{x} \right) = 0 + 1 = 1 \]

### Step 4: Identify the vertical asymptote

Since the function \( y = x + \frac{\sin x}{x} \) is defined for all \( x \) (including \( x = 0 \)), there are no vertical asymptotes.

### Conclusion

The asymptotes of the curve \( y = x + \frac{\sin x}{x} \) are:

1. An oblique asymptote: \( y = x \) as \( x \to \infty \)
2. No vertical asymptotes.