ONE MORE STEP TOWARD BETTER TOMORROW : chapter 15 Worked Out Examples 6

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.

Saturday, 9 August 2014

chapter 15 Worked Out Examples 6

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