Special Limits Involving sin(x), x, and tan(x)
Key Questions

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# . 
Answer
By using:#lim_{x to 0}{sinx}/{x}=1# ,#lim_{x to 0}{tanx}/x=1# .Explanation
Let us look at some details.
Since#tanx={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 functions#y=sin x# and#y=x# are very similar when#x# is near#0# .)
Questions
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Limits Involving Trigonometric Functions

Intuitive Approach to the derivative of y=sin(x)

Derivative Rules for y=cos(x) and y=tan(x)

Differentiating sin(x) from First Principles

Special Limits Involving sin(x), x, and tan(x)

Graphical Relationship Between sin(x), x, and tan(x), using Radian Measure

Derivatives of y=sec(x), y=cot(x), y= csc(x)

Differentiating Inverse Trigonometric Functions