lim sinx / xx -> 0
1
To solve this limit, we can use L’Hôpital’s rule. Applying the rule, we take the derivative of both the numerator and denominator with respect to x:
lim (cosx / x – sinx / x^2) as x -> 0
Then, we can plug in our limit value of x = 0:
lim (cos0 / 0 – sin0 / 0^2)
This simplifies to:
lim (1/0 – 0/0)
Since we ended up with an indeterminate form of 1/0, this limit does not exist.
However, there is a shortcut to this problem if we know the result of another limit. We can use the squeeze theorem along with the identity that sinx <= x <= tanx whenever x is between 0 and pi/2. Thus: 0 <= abs(sin x) / x <= 1/cosx Taking the limits as x -> 0 for each of the expressions above:
0 <= limit as x->0 of abs(sin x) / x <= 1 By the squeeze theorem, we know that the limit of abs(sin x) / x as x -> 0 is equal to 1.
Therefore, by substitution, we have:
limit sinx / x as x -> 0 = limit [(abs(sin x) / x) * (sin x / abs(sin x))] as x -> 0 = 1 * 1 = 1
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