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Examples of calculus

August 16, 2020
Category: Examples of calculus

Examples :

1 Find : $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{\sqrt[3]{1+x}-1}{x}$

Soln. $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{\sqrt[3]{1+x}-1}{x} \left[\frac{0}{0}\mathrm{\ \ form} \right]$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{\frac{d}{dx}\left[\sqrt[3]{1+x}-1\right]}{\frac{d}{dx}(x)}$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{\frac{1}{3}(1+x)^{-\frac{2}{3}}}{1}$

= $\frac{\frac{1}{3}(1+0)^{-\frac{2}{3}}}{1}$

= $\frac{1}{3}$

2 Find : $\begin{matrix}lim\\n\rightarrow \infty \\\end{matrix}$ $\left(\frac{1^2}{n^3}+\frac{2^2}{n^3}+\frac{3^2}{n^3}+.....+\frac{n^2}{n^3}\right)$

Soln. $\begin{matrix}lim\\n\rightarrow \infty \\\end{matrix}$ $\left(\frac{1^2}{n^3}+\frac{2^2}{n^3}+\frac{3^2}{n^3}+.....+\frac{n^2}{n^3}\right)$

= $\begin{matrix}lim\\n\rightarrow \infty \\\end{matrix}$ $\left(\frac{1^2+2^2+3^2+.....+n^2}{n^3}\right)$

= $\begin{matrix}lim\\n\rightarrow \infty \\\end{matrix}$ $\left[\frac{n(n+1)(2n+1)}{6n^3}\right]$ $\left[\frac{\infty}{\infty}\mathrm{\ \ form} \right]$

= $\begin{matrix}lim\\n\rightarrow \infty \\\end{matrix}$ $\left[\frac{(n+1)(2n+1)}{6n^2}\right]$ = $\begin{matrix}lim\\n\rightarrow \infty \\\end{matrix}$ $\left[\frac{2(n+1)+2n+1}{2n}\right]$ [ after differentiating ]

= $\frac{1}{6}$ $\begin{matrix}lim\\n\rightarrow \infty \\\end{matrix}$ $\left[\frac{4n+3}{2n}\right]$ = $\frac{1}{6}$ $\begin{matrix}lim\\n\rightarrow \infty \\\end{matrix}$ $\left[\frac{4}{2}\right]$ [ Again differentiating ]

= $\frac{1}{6}\times 2$

= $\frac{1}{3}$

3 Find : $\begin{matrix}lim\\x\rightarrow 1\\\end{matrix}$ $sec\frac{\pi}{2^x}$ ln x

Soln. $\begin{matrix}lim\\x\rightarrow 1\\\end{matrix}$ $sec\frac{\pi}{2^x}$ ln x [ ∞ × 0 form ] ( Where, ln x = log_ex )

$\begin{matrix}lim\\x\rightarrow 1\\\end{matrix}$ $sec\frac{\pi}{2^x}$ ln x [ ∞ × 0 form ] ( Where, ln x = logx )

= $\frac{ln{x}}{cos{\frac{\pi}{2^x}}}$ [ converted to $\frac{0}{0}$ form ]

= $\frac{\frac{1}{x}}{-sin{\frac{\pi}{2^x}}\cdot\frac{\pi}{2^x}(-log{2})}$

= $\frac{1}{-sin{\frac{\pi}{2}}\cdot\frac{\pi}{2}(-log{2})}$

= $\frac{2}{\pi l o g{2}}$

4 Find : $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\left(\frac{1}{x}-\mathrm{cosec\ } x\right)$

Soln. $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\left(\frac{1}{x}-\mathrm{cosec\ } x\right)$ (∞– ∞ form )

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{sin{x}-x}{xsin{x}}$ [ converted to $\frac{0}{0}$ form ]

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\left(\frac{cos{x}-1}{xcos{x}+sin{x}}\right) \left[\frac{0}{0}\mathrm{\ \ form} \right]$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{-sin{x}}{-xsin{x}+cos{x}+cos{x}}$

= $\frac{0}{0+1+1}$

= 0

5 Find : ( cos x ) $\frac{1}{x^2}$

Soln. ( cos x ) $\frac{1}{x^2}$ [ 1∞] form

Let, A = ( cos x ) $\frac{1}{x^2}$

Then log A = $\frac{logcos{x}}{x^2}$ [ converted to $\frac{0}{0}$ form ]

= $\frac{-tan{x}}{2x}$

= $\frac{-\mathrm{se}\mathrm{c}^2x}{2}$

= – $\frac{1}{2}$

∴ A = (e) $\frac{-1}{2}$ or ( cos x ) $\frac{1}{x^2}$

= e $\frac{-1}{2}$

= $\frac{1}{\sqrt e}$

6 Find : ( |x| )^{sin x}

Soln. ( |x| )^{sin x} [ 0 form ]

Let A = ( |x| )^{sin x}

Then log A = $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ sin x log|x| = $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{log{\left|x\right|}}{\mathrm{cosec\ }x} \left(\frac{-\infty}{\infty}\right)$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{\frac{1}{x}}{\mathrm{cosec\ \ x\ \ cot\ \ }x}$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\left[-\frac{sin{x} \ tan{x}}{x}\right]$

= $\left[-\frac{cos{x} \ \ tan{x}+sin{x} \ \ \mathrm{se} \mathrm{c}^2x}{1}\right]$

= 0

∴ A = e⁰ = 1 or $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ (|x|)^{sin x} = 0

7 Find : $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ ( sec x )^{cot x}

Soln. $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ ( sec x )^{cot x} [ $\infty^0$ orm ]

Let A = $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ ( sec x )^{cot x}

Then log A = $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ cot x log sec x = $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ $\frac{logsec{x}}{tan{x}} \left[\frac{\infty}{\infty}\mathrm{\ \ form} \right]$

= $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ $\frac{\frac{1}{sec{x}} \ sec{x} \ tan{x}}{{sec}^2{x}}$

= $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ $\frac{tan{x}}{{sec}^2{x}} \left[\frac{\infty}{\infty}\mathrm{\ \ form} \right]$

= $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ $\frac{{sec}^2{x}}{2{sec}^2{x} \ tan{x}}$

= $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ $\frac{1}{2tan{x}}$

= $\frac{1}{\infty}$

= 0

∴ A = e⁰ = 1 or $\begin{matrix}lim\\x\rightarrow \frac{\pi}{2}\\\end{matrix}$ ( sec x )^{cot x} = 1

8 Which of the following limits is correct

(a) $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{tan{x}-sin{x}}{{tan}^3{x}}$ = $\frac{1}{6}$

(b) $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{cos{7}x-cos{9}x}{cos{3}x-cos{4}x} = \frac{16}{7}$

(c) $\begin{matrix}lim\\\theta \rightarrow 0\\\end{matrix}$ \frac{1-cos{\theta}}{{sin}^2{2}\theta} = \frac{1}{2}

(d) $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{sin{7}x-sin{x}}{sin{4}x} = \frac{3}{2}$

Soln. (a) We find that at x = 0 the given function takes the form $\frac{0}{0}$ which is an indeterminate form.

The given limit = $\frac{tan{x}-sin{x}}{{tan}^3{x}}$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{sin{x}\left(\frac{1}{cos{x}}-1\right)}{{tan}^3{x}}$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{sin{x}\cdot(1-cos{x})}{{tan}^3{x}\cdot c o s{x}}$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{sin{x}\cdot2{sin}^2{\frac{x}{2}}}{{tan}^3{x}\cdot c o s{x}}$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $2 . \frac{\frac{sin{x}}{x}\cdot\left(\frac{sin{x}/2}{x/2}\right)}{\left(\frac{tan{x}}{x}\right)^3\cdot c o s{x}} . \frac{x\left(\frac{x}{2}\right)^2}{x^3}$

= 2 ^. $\frac{\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}\frac{sin{x}}{x}\cdot\left(\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}\frac{sin{x}/2}{x/2}\right)^2}{\left(\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}\frac{tan{x}}{x}\right)^3} . \frac{1}{\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}cos{x}} . \frac{1}{4}$

= 2

(b)At x = 0 the given function takes the form \frac{0}{0} which is an indeterminate form.

$\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{cos{7}x-cos{9}x}{cos{3}x-cos{4}x}$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{7sin{7}x-9sin{9}x}{3sin{3}x-4sin{4}x}$ ( ∵ Using La’ Hospital Rule ) & $\left[\frac{0}{0}\mathrm{\ \ form} \right]$

= $\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{49cos{7}x-81cos{9}x}{9cos{3}x-16cos{4}x}$ ( ∵ Using La’ Hospital Rule )

= $\frac{32}{7}$

The given limit = $\begin{matrix}lim\\\theta\rightarrow 0\\\end{matrix}$ $\frac{1-cos{\theta}}{{sin}^2{2}\theta}$

= $\begin{matrix}lim\\\theta\rightarrow 0\\\end{matrix}$ $\frac{2{sin}^2\ {\frac{1}{2}}\ \theta}{{sin}^2\ {2}\ \theta}$

= $\begin{matrix}lim\\\theta\rightarrow 0\\\end{matrix}$ $2 . \frac{\left(\frac{sin{\frac{1}{2}}\theta}{\frac{1}{2}\theta}\right)^2\times\left(\frac{1}{2}\theta\right)^2}{\left(\frac{sin{2}\theta}{2\theta}\right)^2\times(2\theta)^2}$

= $2\left(\begin{matrix}lim\\\theta\rightarrow 0\\\end{matrix}\frac{sin{\frac{1}{2}}\theta}{\frac{1}{2}\theta}\right)^2 . \frac{1}{\left(\begin{matrix}lim\\\theta\rightarrow 0\\\end{matrix}\frac{sin{2}\theta}{2\theta}\right)^2} . \frac{1}{16}$

= $2 . 1 . \frac{1}{1^2} . \frac{1}{16}$

= $\frac{1}{8}$

(d) Here we observe that the given function tends to indeterminate form \frac{0}{0} as x → 0.

Now, the given limit

$\begin{matrix}lim\\x\rightarrow 0\\\end{matrix}$ $\frac{sin{7}x-sin{x}}{sin{4}x} = \frac{3}{2}$ ( ∵ Using La’ Hospital Rule )

9. The value of $\frac{xsin{\alpha}-\alpha sin{x}}{x-\alpha}$ is

(a) sin α – α cosα

(b) sin α – αcosα

(c) αsin α – cosα

(d) αsin α + cosα

Soln. [ At x = α the given function takes the form $\frac{0}{0}$ which is an indeterminate form. In similar sums if x → a ≠ 0 then convert the given function using the formula

f (x) = f (a + h ), see limits by substitution. ]

Let x = α; ∴h → 0 when x → α.

∴ the given expression = $\begin{matrix}lim\\x\rightarrow \alpha\\\end{matrix}$ $\frac{xsin{\alpha}-\alpha s i n{\alpha}}{x-\alpha}$

= $\begin{matrix}lim\\x\rightarrow \alpha\\\end{matrix}$ $\frac{(\alpha+h)sin{\alpha}-\alpha s i n{(}\alpha+h)}{\alpha+h-\alpha}$

= $\begin{matrix}lim\\h\rightarrow 0\\\end{matrix}$ $\left[\frac{\alpha\left\{sin{\alpha}-sin{(}\alpha+h)\right\}}{h}+sin{\alpha}\right]$

[ At x = the given function takes the form $\frac{0}{0}$ which is an indeterminate form. In similar sums if x → a ≠ 0 then convert the given function using the formula

f (x) = f (a + h ), see limits by substitution. ]

Let x = α; ∴h → 0 when x → α.

∴the given expression = $\frac{xsin{\alpha}-\alpha s i n{\alpha}}{x-\alpha}$

= $\frac{(\alpha+h)sin{\alpha}-\alpha s i n{(}\alpha+h)}{\alpha+h-\alpha}$

= $\begin{matrix}lim\\h\rightarrow 0\\\end{matrix}$ $\left[\frac{\alpha\left\{sin{\alpha}-sin{(}\alpha+h)\right\}}{h}+sin{\alpha}\right]$

= $\begin{matrix}lim\\h\rightarrow 0\\\end{matrix}$ $\left[\alpha\cdot\frac{2cos{\frac{1}{2}}(\alpha+\alpha+h)\cdot s i n{\frac{1}{2}}\left\{\alpha-(\alpha+h)\right\}}{h}+sin{\alpha}\right]$

= $\begin{matrix}lim\\h\rightarrow 0\\\end{matrix}$ $\left[\alpha\cdot\frac{cos{\left(\alpha+\frac{1}{2}h\right)}\cdot s i n{\left(-\frac{1}{2}h\right)}}{\frac{1}{2}h}+sin{\alpha}\right]$

= $\begin{matrix}lim\\h\rightarrow 0\\\end{matrix}$ $\left[\alpha\cdot c o s{\left(\alpha+\frac{h}{2}\right)}\cdot\frac{-sin{\frac{h}{2}}}{\frac{h}{2}}\right] + sin \alpha$

= αcosα^. ( – 1 ) + sinα

= sinα – cosα

ALTERNATIVE METHOD

Using L’Hospital Rule

$\begin{matrix}lim\\x\rightarrow \alpha\\\end{matrix}$ $\frac{xsin{\alpha}-\alpha s i n{x}}{x-\alpha}$

= $\begin{matrix}lim\\x\rightarrow \alpha\\\end{matrix}$ $\frac{sin{\alpha}(1)-\alpha c o s{x}}{1}$

= sinα – cosα

10 The limit $\begin{matrix}lim\\x\rightarrow \alpha\\\end{matrix}$ $\left(\frac{1^2}{n^3}+\frac{2^2}{n^3}+\frac{3^2}{n^3}+.....+\frac{n^2}{n^3}\right)$ =

(a) $\frac{1}{6}$ (b) $\frac{1}{3}$ (c) $\frac{2}{3}$ (d) None of these

Soln. In solving such problems first find out the sum and then the limit should be taken

Here $\frac{1^2}{n^3} + \frac{2^2}{n^3} + \frac{3^2}{n^3} + ... + \frac{n^2}{n^3}$

= $\left(\frac{1^2+2^2+3^2+.....+n^2}{n^3}\right)$

= $\frac{\frac{n(n+1)(2n+1)}{6}}{n^3}$ , from formula of Algebra = $\frac{1}{n^3}.\frac{2n^3+3n^2+n}{6}.$

$\therefore \begin{matrix}lim\\x\rightarrow \alpha\\\end{matrix}$ $\left(\frac{1^2}{n^3}+\frac{2^2}{n^3}+\frac{3^2}{n^3}+.....+\frac{n^2}{n^3}\right)$

= $\begin{matrix}lim\\x\rightarrow \alpha\\\end{matrix}$ $\frac{1}{n^3}.\frac{2n^3+3n^2+n}{6}$

In solving such problems first find out the sum and then the limit should be taken

Here $\frac{1^2}{n^3} + \frac{2^2}{n^3} + \frac{3^2}{n^3} + ... + \frac{n^2}{n^3}$

= $\left(\frac{1^2+2^2+3^2+.....+n^2}{n^3}\right)$

= $\frac{\frac{n(n+1)(2n+1)}{6}}{n^3}$ , from formula of Algebra = $\frac{1}{n^3}.\frac{2n^3+3n^2+n}{6}.$

∴ $\begin{matrix}lim\\n\rightarrow \alpha\\\end{matrix}$ $\left(\frac{1^2}{n^3}+\frac{2^2}{n^3}+\frac{3^2}{n^3}+.....+\frac{n^2}{n^3}\right)$

= $\begin{matrix}lim\\n\rightarrow \alpha\\\end{matrix}$ $\frac{1}{n^3}.\frac{2n^3+3n^2+n}{6}$

= $\begin{matrix}lim\\n\rightarrow \alpha\\\end{matrix}$ $\left(\frac{2+\frac{3}{n}+\frac{1}{n^2}}{6}\right)$

= $\frac{2+0+0}{6}$

= $\frac{1}{3}$

Differentiation

I $\frac{d}{dx}$ f (x) = (x) = $\begin{matrix}lim\\t\rightarrow x\\\end{matrix} \ \ \frac{f(t)-f(x)}{t-x}$

= $\begin{matrix}lim\\h\rightarrow0\\\end{matrix} \frac{f(x+h)-f(x)}{h}$

II If f : ( a + b ) → R is defined and differentiable on ∀ x ∈ ( a + b ), then f is continuous on x. But its converse is not true.

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