What is limit when x approaches 0 of sinx/x?
Please solve without using L'Hopital's rule.
To find the limit when x approaches 0 of sin(x)/x, we can use the Squeeze Theorem or the fact that the sine function is bounded.
First, note that sin(x) is a bounded function. This means that there exists a positive number M such that |sin(x)| ≤ M for all x.
Now, consider the function f(x) = x. Since x approaches 0 as x approaches 0, we know that lim(x→0) f(x) = 0.
Because sin(x) is bounded and |sin(x)| ≤ M for all x, we can write the following inequality:
-M ≤ sin(x) ≤ M.
Dividing each term by x (assuming x ≠ 0), we get:
-M/x ≤ sin(x)/x ≤ M/x.
Now, let's take the limit as x approaches 0 on both sides of the inequality:
lim(x→0) (-M/x) ≤ lim(x→0) (sin(x)/x) ≤ lim(x→0) (M/x).
As x approaches 0, -M/x approaches negative infinity, and M/x approaches positive infinity.
Therefore, we have:
-∞ ≤ lim(x→0) (sin(x)/x) ≤ ∞.
Since the limit is bounded between -∞ and ∞, we can conclude that the limit when x approaches 0 of sin(x)/x does not exist.