TG TUTORIALS

TG TUTORIALS
TARUN GEHLOT

Saturday, 9 August 2014

chapter 15 Worked Out Examples 6

   Example: 8  

Evaluate the following limits:
(a) \mathop {\lim }\limits_{x \to 0} \sin \dfrac{1}{x}
(b) \mathop {\lim }\limits_{x \to 0} x\sin \dfrac{1}{x}
(c) \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{\ln x}}{x}
(d) \mathop {\lim }\limits_{x \to {0^ + }} x\ln x
(e) \mathop {\lim }\limits_{x \to  + \infty } \dfrac{{{x^n}}}{{n!}}
Solution: 8-(a)

Notice that as x \to 0\dfrac{1}{x} \to \infty , that is , \dfrac{1}{x} has no particular limit to which it converges. Hence \sin \dfrac{1}{x} keeps oscillating between +1 and  - 1  as x becomes smaller and smaller, i.e., x \to 0 Therefore, the limit for this function does not exist.
This is also clear from the graph (approximate) of \sin \dfrac{1}{x} sketched below:
sine of a reciprocal function
Solution: 8-(b)

In this limit, in addition to ,\sin \dfrac{1}{x} ‘x’ is also present. Thus, although \sin \dfrac{1}{x}remains oscillating and does not approach any particular limit, it nevertheless remains somewhere between +1  and  - 1, and when it gets multiplied by x (where x \to 0), the whole product gets infinitesimally small.
That is
\mathop {\lim }\limits_{x \to 0} x\sin \dfrac{1}{x} = 0

graph of a sine functionAgain, this is evident from the graph below:
Solution: 8-(c)

This limit can be evaluated purely by observation as follow:
Although ln x   and x are both tending to infinity, increases very slowly as compared to x.
For example, when x = {e^{10}} , \ln x is just 10. When x = {e^{10000}} (a very large number indeed !), ln x  is just 10000.
Therefore, \dfrac{{\ln x}}{x} decreases and becomes infinitesimally small as x \to \infty , i.e.,
\mathop {\lim }\limits_{x \to \infty } \dfrac{{\ln x}}{x} = 0
(We can also use the LH rule to evaluate the limit above: this rule will be discussed later)
Solution: 8-(d)

Consider x  \ln x.
As x \to {0^ + }\ln x \to  - \infty , so that this limit is of the indeterminate form 0 \times \infty .
But as in parts (b) and (c), try to see that the product becomes infinitesimally small as x \to 0.
For example, at x = {e^{ - 10}}{\rm{ln x  =   - 10 }} and x\ln x = \dfrac{{ - 10}}{{{e^{10}}}} =
At x = {e^{ - 1000}}x\ln x = \dfrac{{ - 1000}}{{{e^{1000}}}} (which is very very small)

Hence, here again,
\mathop {\lim }\limits_{x \to {0^ + }} x\ln x = 0
Solution: 8-(e)

If \left| x \right| < 1, then as n \to \infty  , Num \to 0 and Den \to \infty , so that the limit is 0.
For \left| x \right| = 1, also, the limit is obviously 0.
For \left| x \right| > 1 we write \dfrac{{{x^n}}}{{n!}} as
\dfrac{{{x^n}}}{{n!}} = \dfrac{x}{1} \cdot \dfrac{x}{2} \cdot \dfrac{x}{3}\ldots\dfrac{x}{n}
Now, since x is finite, let N  be the integer just less than or equal to {\rm{x;N = [x]}}
Hence,
\dfrac{{{x^n}}}{{n!}} = \dfrac{x}{1} \cdot \dfrac{x}{2}\ldots\dfrac{x}{N} \cdot \dfrac{x}{{N + 1}} \cdot \dfrac{x}{{N + 2}}\ldots\dfrac{x}{n}
The product of the first N  terms is finite; let it be equal to P.
Thus
\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x^n}}}{{n!}} = P\mathop {\lim }\limits_{n \to \infty } \left\{ {\dfrac{x}{{N + 1}} \cdot \dfrac{x}{{N + 2}}\ldots\dfrac{x}{n}} \right\}
The product inside the limit consists of all terms less than 1. Also successive terms become smaller and smaller and tend to 0 as n \to \infty .
Therefore, this product tends to  0  and hence the value of the overall limit is {\rm{P}} \times {\rm{0 = 0}}
\mathop {\lim }\limits_{n \to \infty } \dfrac{{{x^n}}}{{n!}} = 0
Note: As we mentioned earlier, once we have studied differentiation, we’ll study the L’Hospital’s rule for evaluation of limits of the form \dfrac{0}{0}\;{\rm{or}}\dfrac{\infty }{\infty }. However, it might be useful to know the rule right away – so we provide a brief idea here:
As x \to a if f(x) and g(x) both tend to 0or both tend to infinity, then
\mathop {\lim }\limits_{x \to a} \dfrac{{f\left( x \right)}}{{g\left( x \right)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'\left( x \right)}}{{g'\left( x \right)}} if the latter limit exists
Here are two examples:
(i) \mathop {\lim }\limits_{x \to 0} \dfrac{{\tan x - x}}{{{x^3}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{{{\sec }^2}x - 1}}{{3{x^2}}} \left( {{\rm{still}}\dfrac{0}{0}} \right)
 = \mathop {\lim }\limits_{x \to 0} \dfrac{{2{{\sec }^2}x}}{6}\left( {\dfrac{{\tan x}}{x}} \right)
 = \dfrac{2}{6} = \dfrac{1}{3}
(ii) \mathop {\lim }\limits_{x \to 0} \dfrac{{\sin \left( {\pi {{\cos }^2}x} \right)}}{{{x^2}}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\cos \left( {\pi {{\cos }^2}x} \right)}}{2}\begin{array}{*{20}{c}}  { \times  - 2\pi \cos x}\\  {}  \end{array}\left( {\dfrac{{\sin x}}{x}} \right)
 =  - \dfrac{1}{2} \times  - 2\pi
 = \pi
This rule is simple yet extremely powerful, and in general, you’ll be able to solve most limits using this rule.