Special Limits Involving sin(x), x, and tan(x)
Key Questions
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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 besinx , andg(x) would bex .Thus,
lim_(x->0) (sinx)/x = lim_(x->0) (cosx)/(1) Quite clearly, this limit evaluates to
1 , sincecos 0 is equal to1 . -
Answer
By using:lim_{x to 0}{sinx}/{x}=1 ,lim_{x to 0}{tanx}/x=1 .Explanation
Let us look at some details.
Sincetanx={sinx}/{cosx} ,lim_{x to 0}{tanx}/x = lim_{x to 0}{sinx}/{x}cdot1/{cosx} by the Product Rule,
=(lim_{x to 0}{sinx}/x)cdot(lim_{x to 0}1/{cosx}) by
lim_{x to 0}{sinx}/{x}=1 ,=1cdot1/{cos(0)}=1 -
One of some useful limits involving
y=sin x is
lim_{x to 0}{sin x}/{x}=1 .
(Note: This limit indicates that two functionsy=sin x andy=x are very similar whenx is near0 .)
Questions
Differentiating Trigonometric Functions
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Limits Involving Trigonometric Functions
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Intuitive Approach to the derivative of y=sin(x)
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Derivative Rules for y=cos(x) and y=tan(x)
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Differentiating sin(x) from First Principles
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Special Limits Involving sin(x), x, and tan(x)
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Graphical Relationship Between sin(x), x, and tan(x), using Radian Measure
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Derivatives of y=sec(x), y=cot(x), y= csc(x)
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Differentiating Inverse Trigonometric Functions