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teng125
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may i know is it lim x to 0 (1/sinx - 1/x ) is equals to infinity??
Just to avoid confusion: it has to be 0/0 or ∞/∞ as I said, since there are more indeterminate forms possible. Many can be reduced to one of the two 'allowed' ones to use the rule though.daveb said:To apply L'Hospital's Rule, put the expression in theform f(x)/g(x). If the form is indeterminate, then you can apply the Rule.
This notation represents the limit of a function as its input approaches 0. In this case, the function is (1/sinx - 1/x), which is equal to infinity when x approaches 0.
A limit is a mathematical concept that represents the value that a function approaches as its input gets closer and closer to a particular value. In this case, as x approaches 0, the function (1/sinx - 1/x) gets larger and larger, eventually reaching infinity.
The function (1/sinx - 1/x) is undefined at x=0, as both the numerator and denominator become 0. However, as x gets closer and closer to 0, the denominator becomes smaller and smaller, while the numerator remains constant. This results in a larger and larger value for the function, ultimately approaching infinity.
No, the limit of a function can only be calculated analytically by finding its algebraic expression and evaluating it. Since the function (1/sinx - 1/x) is undefined at x=0, it cannot be evaluated numerically at that point.
The limit of a function can provide information about the behavior of that function near a particular point. In this case, as x approaches 0, the function (1/sinx - 1/x) increases without bound, indicating that the function is discontinuous at x=0. This can be useful in understanding the properties of the function and its graph.