How do you find the limit #lim_(x->0)sin(x)/x# ?

Answer 1

We will use l'Hôpital's Rule.

l'Hôpital's rule states:

#lim_(x->a) f(x)/g(x) = lim_(x->a) (f'(x))/(g'(x))#
In this example, #f(x)# would be #sinx#, and #g(x)# would be #x#.

Thus,

#lim_(x->0) (sinx)/x = lim_(x->0) (cosx)/(1)#
Quite clearly, this limit evaluates to #1#, since #cos 0# is equal to #1#.
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Answer 2

To find the limit of ( \lim_{x \to 0} \frac{\sin(x)}{x} ), you can use L'Hôpital's Rule or the Maclaurin series expansion of ( \sin(x) ). Applying L'Hôpital's Rule yields ( \lim_{x \to 0} \frac{\cos(x)}{1} ), which evaluates to ( \cos(0) = 1 ). Alternatively, you can use the Maclaurin series expansion of ( \sin(x) ) to get ( \lim_{x \to 0} \frac{x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots}{x} ), simplifying to ( \lim_{x \to 0} \left(1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \ldots\right) ), which equals ( 1 ).

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Answer from HIX Tutor

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.

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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