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## 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:
 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
 $\mathop {\lim }\limits_{x \to 0} x\sin \dfrac{1}{x} = 0$

Again, 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 $0$or 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.