if limit x tends to infinity

Then find

(Cosx+sinx)/x^2 please answer with solution

We need to evaluate the limit:

lim (x → ∞) (cos x + sin x) / x²

Step 1: Understanding the given function
The numerator of the function is cos x + sin x, while the denominator is x².

The trigonometric functions cos x and sin x are oscillatory functions with values always between -1 and 1.
The denominator x² grows infinitely large as x → ∞.
Step 2: Finding the bounds for the numerator
Since cos x and sin x both range between -1 and 1, their sum also satisfies the inequality:

-1 - 1 ≤ cos x + sin x ≤ 1 + 1
-2 ≤ cos x + sin x ≤ 2

This means that the numerator is always within [-2, 2].

Step 3: Dividing by x²
Now, we divide the entire inequality by x²:

-2 / x² ≤ (cos x + sin x) / x² ≤ 2 / x²

Since x² → ∞ as x → ∞, we observe that:

-2 / x² → 0 and 2 / x² → 0

Thus, by the Squeeze Theorem, we conclude that:

lim (x → ∞) (cos x + sin x) / x² = 0.

Final Answer:
The required limit is 0.