prove lim x->0 sinx/x=1

To prove the limit as x approaches 0 of sin(x)/x is equal to 1, we can use the concept of trigonometric limits.

By using a Trigonometric Identity, we have sin(x)/x = 1/(x/sin(x)).

Now, let's focus on the denominator (x/sin(x)).

As x approaches 0, sin(x) also approaches 0, because sine function is continuous at x=0.

Therefore, we can rewrite the denominator as (x/sin(x)) as (0/0), which is an indeterminate form.

To resolve this indeterminate form, we can use L'Hopital's Rule, which allows us to differentiate numerator and denominator separately.

When we differentiate the numerator and denominator, we get:

d/dx (x) = 1, and d/dx (sin(x)) = cos(x).

Now, we can rewrite the expression as 1/(x/sin(x)), which is equivalent to 1/(1/cos(x)) = cos(x).

So, the limit of (x/sin(x)) as x approaches 0 is equal to cos(0) = 1.

Since we've shown that the denominator approaches 1 and the numerator is 1, we can conclude that the limit as x approaches 0 of sin(x)/x is indeed equal to 1.

Therefore, we have proven that lim x->0 sin(x)/x = 1.