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How do you find #lim sinx/x# as #x->0# using l'Hospital's Rule?

Answer 1

Ezekiel Alexander

#lim_(x->0) sinx/x =1#

#lim_(x->0) sinx/x# is a well known standard limit #=1#.

Here, we are asked to use l'Hospital's rule since the limit reduces to the indeterminate form #0/0#

Thus, #lim_(x->0) sinx/x = lim_(x->0) (d/dx sinx)/(d/dx x)#

#= lim_(x->0) cosx/1 = 1/1 =1#

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Answer 2

Evelyn Agard

To find (\lim_{x \to 0} \frac{\sin x}{x}) using L'Hospital's Rule, we can apply the rule which states that if the limit of the ratio of two functions is of the form (\frac{0}{0}) or (\frac{\infty}{\infty}), then the limit of the ratio is equal to the limit of the ratio of their derivatives, provided the latter limit exists.

Given (\lim_{x \to 0} \frac{\sin x}{x}), we have (\frac{0}{0}) as (x \to 0). So, we can differentiate the numerator and the denominator and find their limits as (x \to 0).

[ \lim_{x \to 0} \frac{\sin x}{x} = \lim_{x \to 0} \frac{\frac{d}{dx}(\sin x)}{\frac{d}{dx}(x)} ]

[ = \lim_{x \to 0} \frac{\cos x}{1} ]

[ = \cos 0 ]

[ = 1 ]

So, (\lim_{x \to 0} \frac{\sin x}{x} = 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.

How do you find #lim sinx/x# as #x-&gt;0# using l'Hospital's Rule?

How do you find #lim sinx/x# as #x->0# using l'Hospital's Rule?