#implies f"(x)=lim_(x khổng lồ 0) 1/(sinx/x)=(lim_(x lớn 0)1)/(lim_(x lớn 0)(sinx/x))=1/1=1#


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We need to know the important trigonometric limit:

#lim_(xrarr0)sinx/x=1#

So, we see that:

#lim_(xrarr0)x/sinx=lim_(xrarr0)(sinx/x)^-1=(lim_(xrarr0)sinx/x)^-1=1^-1=1#


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If we try to calculate the limit directly, we can see that is an indeterminate form:

#lim_x rarr 0 x/sin x = 0/sin 0 = 0/0 ?#

To solve it, we can apply the L"Hôpital"s rule:

Given two functions #f# and #g# differentiable at #x = a#, it holds that:

If #lim_x rarr a f (x)/g (x) = 0/0# or #lim_x rarr a f (x)/g (x) = oo/oo# then:

#lim_x rarr a f (x)/g (x) = lim_x rarr a f" (x)/g" (x)#

So we have:

#lim_x rarr 0 x/sin x = lim_x rarr 0 1/cos x = 1/cos 0 = 1/1 = 1#.

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NOTEThe question was posted in "Determining Limits Algebraically" , so the use of L"Hôpital"s rule is NOT a suitable method lớn solve the problem. Therefore this solution is invalid.

ANSWER to THE NOTEThis limit can not be solved using only algebraic concepts as the function #sin (x)# is not an algebraic function. We may use a Taylor series to lớn approximate #sin (x)# by a polynomial và thus use a linear or quadratic approximation that would calculate the limit, but also emerge the issue of derivatives.

The answer above that uses the limit #lim_x rarr 0 sin x/x# also is invalid (using the criteria indicated by the note) because this limit cited needs also L"Hôpital"s rule to lớn be improved. It is not correct lớn say that is an important limit & that is why we must know if we can not prove it in the context that is intended for use.

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I think the person who wrote the first chú ý confuses the term "algebraic" in the expression "Determining Limits algebraically" in the true meaning of the word algebraic, regardless of that expression usually means "Determination of limits analytically", ie, by calculation and not graphically, numerically (using approximations) or intuitive.