lim x->0 sinx/x
To find the limit of sin(x)/x as x approaches 0, we can use a common trigonometric limit
To find the limit of sin(x)/x as x approaches 0, we can use a common trigonometric limit.
The limit of sin(x)/x as x approaches 0 is a fundamental limit in calculus with various applications. To evaluate this limit, we can apply L’Hopital’s Rule or use a trigonometric identity.
Method 1: Using L’Hopital’s Rule
L’Hopital’s Rule allows us to differentiate the numerator and denominator separately and then take the limit again. Here’s how we can apply L’Hopital’s Rule:
lim x->0 sin(x)/x = lim x->0 cos(x)/1 = cos(0)/1 = 1
In this approach, we differentiate the numerator, which is cos(x), and the denominator, which is 1, and then evaluate the limit.
Method 2: Using Trigonometric Identity
Alternatively, we can use a well-known trigonometric identity involving the sine function:
sin(x) = x – (x^3)/3! + (x^5)/5! – (x^7)/7! + …
Using this identity, we can rewrite sin(x)/x as:
sin(x)/x = (x – (x^3)/3! + (x^5)/5! – (x^7)/7! + …) / x
Now, cancelling out one ‘x’ from the numerator and denominator, we get:
sin(x)/x = 1 – (x^2)/3! + (x^4)/5! – (x^6)/7! + …
As x approaches 0, each of the terms with higher powers of x becomes negligible, and we are left with the first term only, which is 1:
lim x->0 sin(x)/x = 1
So, both methods yield the same result: the limit of sin(x)/x as x approaches 0 is 1.
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