# Examples On Evaluating Limits Set-2

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Example - 3

Find the values of a and b if $$\mathop {\lim }\limits_{x \to \infty } \left\{ {\frac{{{x^2} - 1}}{{x + 1}} - \left( {ax + b} \right)} \right\} = 2$$

Solution: The limit can be rearranged as

$\mathop {\lim }\limits_{x \to \infty } \left\{ {\left( {x - 1} \right) - \left( {ax + b} \right)} \right\}$

$= \mathop {\lim }\limits_{x \to \infty } \left\{ {\left( {1 - a} \right)x - \left( {1 + b} \right)} \right\} = 2$

Since the limit is finite, the coefficient of a has to be necessarily 0.

Therefore

$\begin{gathered} \;\;1 - a = 0\;\;\; \Rightarrow & \;\;a = 1 \\ - 1 - b = 2\;\;\; \Rightarrow & \;\;b = - 3 \\ \end{gathered}$

Example – 4

Evaluate \begin{align}\mathop {\lim }\limits_{n \to \infty } \frac{{\left[ x \right] + \left[ {{2^2}x} \right] + \left[ {{3^2}x} \right] + ... + \left[ {{n^2}x} \right]}}{{{n^3}}}\end{align} where [ . ] represents the greatest integer function.

Solution: One might say that since the terms in the numerator are all integral, the numerator is not continuous and hence the limit will not exist. However note first of all that the limit is on n, and the secondly, addition of a large number of integral terms in the numerator $$\left( {n \to \infty } \right)$$ would tend to ‘overcome’ or ‘make negligible’ the effect of fractional parts that would otherwise have been present had there been no greatest integer functions on any of the terms. This implies that whether I consider $$\left( {\left[ x \right] + \left[ {{2^2}x} \right] + \,\,.......\,\,\left[ {{n^2}x} \right]} \right),$$ or $$\left( {x + {2^2}x + ......{n^2}x} \right)$$ , as n becomes larger and $$\to \infty$$ , the difference between these two terms becomes negligible in comparison to their own magnitude. Hence, the limit in question is equivalent to

\begin{align} &\mathop {\lim }\limits_{n \to \infty } \frac{{x + {2^2}x + {3^2}x + ...{n^2}x}}{{{n^3}}} \\&\quad = \mathop {\lim }\limits_{n \to \infty } \frac{{x\left( {1 + {2^2} + {3^2} + ...{n^2}} \right)}}{{{n^3}}} \\&\quad = \frac{x}{6}\mathop {\lim }\limits_{n \to \infty } \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{{n^3}}}\\&\quad = \frac{x}{6}\mathop {\lim }\limits_{n \to \infty } \left( {1 + \frac{1}{n}} \right) \cdot \left( {2 + \frac{1}{n}} \right)\\&\quad = \frac{x}{3} \\ \end{align}

For students who like more rigor here is the proof of the above result using Sandwich theorem (In this proof, it will become clear that the effect of the fractional part’s is negligible as $$n \to \infty$$

\left. \begin{align} x - 1 < \left[ x \right] \le x \\ {2^2}x - 1 < \left[ {{2^2}x} \right] \le {2^2}x \\ {3^2}x - 1 < \left[ {{3^2}x} \right] \leqslant {3^2}x \\ {n^2}x - 1 < \left[ {{n^2}x} \right] \le {n^2}x \\ \end{align} \right\}By\; definition\; of\; the\; greatest \;integer \;function

Addition of the these inequalities yields

$\left( {x + {2^2}x + {3^2}x + ... + {n^2}x - n} \right) < \left[ x \right] + \left[ {{2^2}x} \right] + ...\left[ {{n^2}x} \right] \leqslant \left( {x + {2^2}x + ...{n^2}x} \right)$

Division by $${n^3}$$ and application of $$\mathop {\lim }\limits_{n \to \infty }$$ on all three terms yields:

$\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}x - n}}{{{n^3}}}} \right\} < \mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{\left[ x \right] + \left[ {{2^2}x} \right] + ...\left[ {{n^2}x} \right]}}{{{n^3}}}} \right\} \le \mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}x}}{{{n^3}}}} \right\}$

It is easy to that the left and right limits are both $$\frac{x}{3}$$ , and hence the centre limit is also $$\frac{x}{3}$$ .

Example - 5

Evaluate the following limits:

 $$\rm{(a)} \quad \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\cos x - \sin x}}{{\left( {\frac{\pi }{4} - x} \right)\left( {\cos x + \sin x} \right)}}$$ $$\rm{(b)}\quad \mathop {\lim }\limits_{x \to a} \frac{{\cos \sqrt x - \cos \sqrt a }}{{x - a}}$$ $$\rm{(c)} \quad \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin x}}{{{x^3}}}$$ $$\rm{(d)}\quad \mathop {\lim }\limits_{n \to \infty } \left\{ {\cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}}...\cos \frac{x}{{{2^n}}}} \right\}$$

Solution : Before proceeding to solve these limits, it should be mentioned that most limits can be evaluated in more than one manner. (In fact, L’Hospital’s rule is a technique that can be used to solve most indeterminate limits: we will discuss it in detail in the topic of differentiation). In these examples and the ones that follow we will be discussing multiple ways for solving limits, wherever they are important.

(a) \begin{align}\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\cos x - \sin x}}{{\left( {\frac{\pi }{4} - x} \right)\left( {\cos x + \sin x} \right)}} = \mathop {\lim }\limits_{x \to \pi /4} \frac{{1 - \tan x}}{{\left( {\frac{\pi }{4} - x} \right)\left( {1 + \tan x} \right)}}\end{align} (Dividing Num and Den by cos x)

Let \begin{align}x = \frac{\pi }{4} + h,\end{align} so that \begin{align}\tan x = \frac{{1 + \tan \,h}}{{1 - \tan \,h}}\end{align} and the limit above becomes

$\mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{1 - \frac{{1 + \tan \,h}}{{1 - \tan \,h}}}}{{ - h}}} \right\} \cdot \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \left\{ {\frac{1}{{1 + \tan \,x}}} \right\}$

$= \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{ - 2\,\tan \,h}}{{ - h\left( {1 - \tan \,h} \right)}}} \right\} \cdot \frac{1}{2} = \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{\tan \,h}}{h}.\frac{1}{{1 - \tan \,h}}} \right\} = 1$

Alternatively, we could proceed by multiplying the numerator and denominator of the original limit by $$\left( {cosx{\rm{\; }} + {\rm{\; }}sinx} \right):$$

\begin{align}&\mathop {\lim }\limits_{x \to \frac{\pi }{4}} \left\{ {\frac{{\cos x - \sin x}}{{\left( {\frac{\pi }{4} - x} \right)\left( {\cos x + \sin x} \right)}} \cdot \frac{{\cos x + \sin x}}{{\cos x + \sin x}}} \right\}\\&= \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{{{\cos }^2}x - {{\sin }^2}x}}{{\left( {\frac{\pi }{4} - x} \right){{\left( {\cos x + \sin x} \right)}^2}}}\\&= \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{{\cos 2x}}{{\left( {\frac{\pi }{4} - x} \right)}}\, \cdot \mathop {\lim }\limits_{x \to \frac{\pi }{4}} \frac{1}{{{{\left( {\cos x + \sin x} \right)}^2}}}\\&= \mathop {\lim }\limits_{h \to 0} \frac{{ - \sin 2h}}{{ - h}} \cdot \frac{1}{2}\;\;\;\;\left( {{\text{we let }}x = \frac{\pi }{4} + h\;{\text{so that}}\cos 2x = \cos \left( {\frac{\pi }{2} + 2h} \right) = - \sin 2\;h} \right)\\&= \mathop {\lim }\limits_{h \to 0} \,\,\frac{{\sin 2h}}{{2h}}\\&= 1\end{align}

(b) \begin{align}\mathop {\lim }\limits_{x \to a} \frac{{\cos \,\sqrt x - \cos \sqrt a }}{{x - a}}\end{align}

$= \mathop {\lim }\limits_{x \to a} \left\{ {\frac{{ - 2\,\sin \left( {\frac{{\sqrt x + \sqrt a }}{2}} \right)\sin \left( {\frac{{\sqrt x - \sqrt a }}{2}} \right)}}{{\left( {\sqrt x + \sqrt a } \right)\left( {\sqrt x - \sqrt a } \right)}}} \right\}$

${\rm{\{We\;used\;}}cosC - cosD = - 2\sin \left( {\frac{{C + D}}{2}} \right)\sin \left( {\frac{{C - D}}{2}} \right)$

$= - \frac{{\sin \sqrt a }}{{2\sqrt a }} \cdot \mathop {\lim }\limits_{x \to a} \left\{ {\frac{{\sin \left( {\frac{{\sqrt x - \sqrt a }}{2}} \right)}}{{\left( {\frac{{\sqrt x - \sqrt a }}{2}} \right)}}} \right\}$

$= - \frac{{\sin \sqrt a }}{{2\sqrt a }}$

(c) This limit can be solved very easily by using the expansion series for $$\text{sin}x:$$

$\sin \,x = x - \frac{{\,{x^3}}}{{3!}} + \frac{{\,{x^5}}}{{5!}} - ...\infty$

\begin{align}\text{Therefore, }\mathop {\lim }\limits_{x \to 0} \frac{{x - \sin \,x}}{{{x^3}}}&= \mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{\frac{{{x^3}}}{{3!}} - \frac{{{x^5}}}{{5!}} + ...\infty }}{{{x^3}}}} \right\}\\ &= \mathop {\lim }\limits_{x \to 0} \;\left\{ {\frac{1}{{3!}} - \frac{{{x^2}}}{{5!}} + \frac{{{x^4}}}{{7!}} - ...\infty } \right\}\\&= \frac{1}{{3!}} = \frac{1}{6}\end{align}

Alternatively, we know that $$\text{sin3}x= 3\,\text{sin}x-4\text{sin}^3x$$

Hence,$$3x{\rm{ }}-{\rm{ }}3{\rm{\; }}sin{\rm{ }}x{\rm{ }} = {\rm{ }}3{\rm{ }}x{\rm{ }}--{\rm{ }}sin{\rm{ }}3x{\rm{ }}-{\rm{ }}4{\rm{ }}si{n^3}x$$

\begin{align}& L = \mathop {\lim }\limits_{x \to 0} \frac{{x - \sin \,x}}{{{x^3}}} = \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \frac{{3x - 3\,\sin \,x}}{{{x^3}}} \\&\quad = \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \left\{ {\frac{{3x - \sin \,3x}}{{{x^3}}} - \frac{{4{{\sin }^3}x}}{{{x^3}}}} \right\}\\ &\quad= \frac{1}{3}\mathop {\lim }\limits_{x \to 0} \left\{ {27 \cdot \frac{{\left( {3x} \right) - \sin \left( {3x} \right)\,}}{{{{\left( {3x} \right)}^3}}} - \frac{{4{{\sin }^3}x}}{{{x^3}}}} \right\} \\ &\quad= \frac{1}{3} \cdot \left\{ {27 \cdot \mathop {\lim }\limits_{\theta \to 0} \frac{{\theta - \sin \theta \,}}{{{\theta ^3}}} - 4\mathop {\lim }\limits_{x \to 0} {{\left( {\frac{{\sin \,x}}{x}} \right)}^3}} \right\} \\ &\quad= 9L - \frac{4}{3} \\ \end{align}

Hence, \begin{align}L = 9L - \frac{4}{3}\;\;\;{\rm{or}}\;\;L = \frac{1}{6}\end{align}

(d) As in Example - 1 Part - (b), we try to reduce this expression into a closed form by multiplying it with an appropriate factor as follows:

\begin{align}&\mathop {\lim }\limits_{n \to \infty } \left\{ {\cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}}...\cos \frac{x}{{{2^n}}} \cdot \frac{{\sin \frac{x}{{{2^n}}}}}{{\sin \frac{x}{{2n}}}}} \right\}\\\\& = \mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{\cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}}...\cos \frac{x}{{{2^{n - 1}}}} \cdot \left( {\overbrace {2\cos \frac{x}{{{2^n}}} \cdot \sin \frac{x}{{{2^n}}}}^{{\rm{combine}}\,:\,\sin \,2\theta = 2\sin \theta \cos \theta }} \right)}}{{2\sin \frac{x}{{{2^n}}}}}} \right\}\\\\& = \mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{\cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}}...\overbrace {\cos \frac{x}{{{2^{n - 1}}}} \cdot \sin \frac{x}{{{2^{n - 1}}}}}^{{\rm{combine}}}}}{{2\sin \frac{x}{{{2^n}}}}}} \right\}\\\\& = \mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{\cos \frac{x}{2} \cdot \cos \frac{x}{{{2^2}}}...\cos \frac{x}{{{2^{n - 2}}}} \cdot \sin \frac{x}{{{2^{n - 2}}}}}}{{{2^2} \cdot \sin \frac{x}{{{2^n}}}}}} \right\}\end{align}

:

:

:

\begin{align}& = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \,x}}{{{2^n}\sin \left( {\frac{x}{{{2^n}}}} \right)}}\\\\& = \mathop {\lim }\limits_{n \to \infty } \frac{{\sin \,x}}{{\frac{{x \cdot \left( {\sin \left( {\frac{x}{{{2^n}}}} \right)} \right)}}{{\left( {\frac{x}{{{2^n}}}} \right)}}}} = \frac{{\sin \,x}}{x} \cdot \frac{1}{{\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{\sin \left( {\frac{x}{{{2^n}}}} \right)}}{{\left( {\frac{x}{{{2^n}}}} \right)}}} \right\}}}\end{align}

Now, as $$n \to \infty ,\frac{x}{{{2^n}}} \to 0$$ and hence

$\mathop {\lim }\limits_{n \to \infty } \left\{ {\frac{{\sin \frac{x}{{{2^n}}}}}{{\frac{x}{{{2^n}}}}}} \right\} = 1$

Therefore, our final result is \begin{align}\frac{{\sin \,x}}{x}\end{align}