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Saturday, 9 August 2014

chapter 5 Some Standard Limits 1

(A) \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = 1;\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = 1
Both the limits above are indeterminate, of the form \dfrac{0}{0}. We are discussing here a geometric interpretation of these limits. Consider a sector OAP of a unit circle as shown in the figure below.
image showing sector of a circle
We see that
{\rm{area}}\,\left( {\Delta OAP} \right) < {\rm{area}}\,\left( {{\,\rm{sector}}\,OAP} \right) < {\rm{area}}\,\left( {\Delta OAB} \right) or
\dfrac{1}{2}\sin x< \dfrac{1}{2}x < \dfrac{1}{2}\tan x or
\dfrac{{\sin x}}{x} < 1 < \dfrac{{\tan x}}{x}
What happens as x decreases or as x \to 0?
We see that the difference between the three areas considered above tends to decrease;
{\rm{area}}\left( {\Delta OAP} \right) \to {\rm{area}}\left( {{\rm{sector}}\,OAP} \right) \leftarrow {\rm{area}}\left( {\Delta OAB} \right)
 \Rightarrow   \sin x \to x\,{\rm{and}}\,x \to \tan x
 \Rightarrow   \dfrac{{\sin x}}{x} \to 1 (but remains less than 1 )
\dfrac{{\tan x}}{x} \to 1 (remains greater than 1 )
 \Rightarrow \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin x}}{x} = {1^ - };\mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x}}{x} = {1^ + }
Consider the limits below and you should understand:It is important to observe how any function approaches a limit. For example, in the case above, as x \to 0,\dfrac{{\sin x}}{x} approaches 1 from the left side while \dfrac{{\tan x}}{x}approaches 1 from the right side. This makes a big difference. Why?
\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\sin x}}{x}} \right] = 0;\mathop {\lim }\limits_{x \to 0} \left[ {\dfrac{{\tan x}}{x}} \right] = 1
(B)\mathop {\lim }\limits_{x \to \infty } {\left( {1 + \dfrac{1}{x}} \right)^x} = e
Consider the expression for f\left( x \right) = {\left( {1 + \dfrac{1}{x}} \right)^x} . As x gets larger and larger or as x \to \infty  , the base \left( {1 + \dfrac{1}{x}} \right) gets closer to 1 while the exponent (x), tends to infinity. Hence, this limit is of the indeterminate form {1^\infty }. Its very important to get a ‘feel’ that the value {\left( {1 + \dfrac{1}{x}} \right)^x} will converge to a fixed, definite value, as xincreases. You should get this feel by looking at the table below.
x1 + \dfrac{1}{x}{\left( {1 + \dfrac{1}{x}} \right)^x}
122
101.12.5937\ldots
1001.012.7048\ldots
10001.0012.7169\ldots
100001.00012.7181\ldots
1000001.000012.7182\ldots
Try to show that the limit of {\left( {1 + \dfrac{1}{x}} \right)^x} is bounded and lies between and 2 and 3, that is, 2 < e < 3 (you can use the binomial theorem in conjunction with the sandwich theorem to prove this, by first proving it for an integral x)We see that as x becomes larger, the term {\left( {1 + \dfrac{1}{x}} \right)^x} converges to some value (This can be proved) This limiting value is denoted by ee is an irrational number and its value is e = 2.7182\ldots.
The limit we have just seen is extremely important and will be widely used subsequently.
Note another important point:
If
f\left( x \right) \to 0 {\rm{as}} x \to a,{\rm{then}}
\mathop {\lim }\limits_{x \to a} {\left( {1 + f(x)} \right)^{\dfrac{1}{{f(x)}}}} = \mathop {\lim }\limits_{f(x) \to 0} {\left( {1 + f\left( x \right)} \right)^{\dfrac{1}{{f(x)}}}}
\mathop {\lim }\limits_{y \to 0} {\left( {1 + y} \right)^{\dfrac{1}{y}}} = e\left( {y = f(x)} \right)
Hence, any limit of this form has the value e.
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