lim x->0 1-cosx/x
0
To solve this limit problem, we can start by using the L’Hopital’s rule, which states that if we have a limit of the form f(x)/g(x) and both f(x) and g(x) approach zero (or infinity) as x approaches some value c, then the limit of f(x)/g(x) as x approaches c is equal to the limit of the derivative of f(x) divided by the derivative of g(x) as x approaches c (provided that the limit of the derivative of g(x) is not zero).
Using this rule, let’s find the derivative of f(x) and g(x):
f(x) = 1 – cos x
f'(x) = sin x
g(x) = x
g'(x) = 1
Now we can rewrite the original limit as:
lim x->0 (1 – cos x)/x
Applying the L’Hopital’s rule, we get:
lim x->0 (sin x)/1
= 0/1
= 0
Therefore, the limit of (1 – cos x)/x as x approaches 0 is equal to 0.
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