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chandubaba
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prove that limit of (sin (x^0)/x) as x tends to zero is π/180(ie pi by 180)
I agree with arild.chandubaba said:prove that limit of (sin (x^0)/x) as x tends to zero is π/180(ie pi by 180)
The limit of sin(x0)/x as x→0 is undefined. This is because the denominator becomes 0, which results in an indeterminate form.
The limit of sin(x0)/x as x→0 is not equal to π/180. This is because the limit does not exist. However, the value of sin(x0)/x as x approaches 0 is equal to π/180, which is a result of the limit approaching this value from both the positive and negative sides.
The limit of sin(x0)/x as x→0 cannot be calculated using traditional limit methods, such as direct substitution or algebraic manipulation. It can only be determined by considering the behavior of the function as x approaches 0 from both the positive and negative sides.
No, the limit of sin(x0)/x as x→0 is not equal to 0. As x approaches 0, the value of sin(x0)/x approaches π/180, not 0. This is because the sine function has a non-zero value at x=0, which affects the overall limit.
No, L'Hopital's rule is not applicable in this case. This is because the limit of sin(x0)/x as x→0 is not in an indeterminate form that can be rewritten as a fraction of two functions, which is a requirement for using L'Hopital's rule.