# What is the limit of # sinx /(x^2 - 4x)# as x approaches 0?

To find the limit of ( \frac{\sin{x}}{x^2 - 4x} ) as ( x ) approaches 0, we can use L'Hôpital's Rule. Taking the derivative of both the numerator and denominator separately:

[ f(x) = \sin{x} ] [ g(x) = x^2 - 4x ]

[ f'(x) = \cos{x} ] [ g'(x) = 2x - 4 ]

Now, applying L'Hôpital's Rule:

[ \lim_{x \to 0} \frac{\sin{x}}{x^2 - 4x} = \lim_{x \to 0} \frac{\cos{x}}{2x - 4} ]

Substituting ( x = 0 ) into the expression:

[ = \frac{\cos{0}}{2(0) - 4} = \frac{1}{-4} = -\frac{1}{4} ]

Therefore, the limit of ( \frac{\sin{x}}{x^2 - 4x} ) as ( x ) approaches 0 is ( -\frac{1}{4} ).

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exists, then you have

In your case, you have

which means that

Therefore

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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