lim x->0 sinx/x
1
To evaluate the limit of sinx/x as x approaches 0, we can apply the squeeze theorem. The inequality:
-1 ≤ sin(x) ≤ 1
holds for all x, since the sine function oscillates between the values of -1 and 1. We can further manipulate this inequality by dividing everything by x:
-1/x ≤ sin(x)/x ≤ 1/x
Note that the inequality still holds, since x is approaching 0 from the left and right sides. Now, taking the limit as x approaches 0:
lim x→0 (-1/x) ≤ lim x→0 (sin(x)/x) ≤ lim x→0 (1/x)
As x approaches 0, the left and right-hand limits both approach negative and positive infinity, respectively. Therefore, we can apply the squeeze theorem to conclude that:
lim x→0 (sin(x)/x) = 1
Hence, the limit of sinx/x as x approaches 0 is equal to 1.
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