lim x->0 sinx/x
To find the limit of sin(x)/x as x approaches 0, we can substitute 0 into the function but that would result in an undefined value, because dividing by zero is not defined
To find the limit of sin(x)/x as x approaches 0, we can substitute 0 into the function but that would result in an undefined value, because dividing by zero is not defined.
However, there is a special property of trigonometric functions and limits called the Squeeze Theorem or the Sandwich Theorem, which can be used to find the limit in this case.
The Squeeze Theorem states that if we have three functions, f(x), g(x), and h(x), such that f(x) ≤ g(x) ≤ h(x) for all x close to a particular value, except possibly at x=a, and both f(x) and h(x) approach the same limit L as x approaches a, then g(x) also approaches L as x approaches a.
In the case of sin(x)/x as x approaches 0, we can use the Squeeze Theorem by finding two functions whose limits we already know.
The first step is to observe that for all x ≠ 0, the inequality |sin(x)| ≤ |x| holds. This simply means that the absolute value of sin(x) is always less than or equal to the absolute value of x.
Hence, we have:
-|x| ≤ sin(x)/x ≤ |x|
As x approaches 0, all three functions -|x|, sin(x)/x, and |x| approach 0.
This means that sin(x)/x is “sandwiched” between -|x| and |x|, and since both -|x| and |x| approach 0, we can conclude that sin(x)/x also approaches 0 as x approaches 0.
Therefore, we can write:
lim(x->0) sin(x)/x = 0
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