Chapter 9 Indeterminate Forms-I

• 00$\frac{0}{0}$
• ∞−∞$\infty -\infty$
• 00$0^0$
• 0×∞$0\times \infty$

• ∞∞$\frac{\infty }{\infty }$
• 1∞$1^{\infty }$
• ∞0${\infty }^0$

are indeterminate forms.

9.1 Exercise 5

9.1.1 Question 1

Evaluate the limits:

1. limx→2x3−2x2+2x−4x2−5x+6${lim}_{x\to 2}\frac{x^3-2x^2+2x-4}{x^2-5x+6}$

This is of form 00$\frac{0}{0}$, so using L’Hospital rule,

=limx→23x2−4x+22x−5=12−8+2−1=−6$$\begin{array}{cc}& =\underset{x\to 2}{lim}\frac{3x^2-4x+2}{2x-5}\\ & =\frac{12-8+2}{-1}\\ & =-6\end{array}$$

2. limx→0tanx−xx−sinx${lim}_{x\to 0}\frac{\tan x-x}{x-\sin x}$

This is also of form 00$\frac{0}{0}$. So applying L’Hospital rule,

=limx→0sec2x−11−cosx⟹(00)=limx→02sec2xtanxsinx=limx→02sec2xsinxsinxcosx=limx→02sec3x=2$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{{\sec }^{2}x-1}{1-\cos x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{2{\sec }^{2}x\tan x}{\sin x}& \\ & =\underset{x\to 0}{lim}\frac{2{\sec }^{2}x\sin x}{\sin x\cos x}=\underset{x\to 0}{lim}2{\sec }^{3}x& \\ & =2& \end{array}$$

3. limx→axn−anx−a${lim}_{x\to a}\frac{x^n-a^n}{x-a}$

This is also of form 00$\frac{0}{0}$. So applying L’Hospital rule,

=limx→anxn−1−01=nan−1$$\begin{array}{cc}& =\underset{x\to a}{lim}\frac{n{x}^{n-1}-0}{1}\\ & =n{a}^{n-1}\end{array}$$

4. limx→0x−sin−1xsin3x${lim}_{x\to 0}\frac{x-{\sin }^{-1}x}{{\sin }^{3}x}$

The expression can be written as, =limx→0x−sin−1xx3(sinxx)3$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{x-{\sin }^{-1}x}{x^3{\left(\frac{\sin x}{x}\right)}^3}\end{array}$$

Using property of multiplication of two limits,

=limx→0x−sin−1xx3×1, as limx→0sinxx=1=limx→0x−sin−1xx3$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{x-{\sin }^{-1}x}{x^3\times 1},\text{as}\underset{x\to 0}{lim}\frac{\sin x}{x}=1\\ & =\underset{x\to 0}{lim}\frac{x-{\sin }^{-1}x}{x^3}\end{array}$$

From the chapter on Taylor and Maclaurin series, we know the expansion of,

sin−1x=x+x33!+9x55!+…$$\begin{array}{cc}{\sin }^{-1}x& =x+\frac{x^3}{3!}+\frac{9x^5}{5!}+\dots \end{array}$$

So,

=limx→0x−sin−1xx3=limx→0x−(x+x33!+9x55!+…)x3=limx→0−x33!−9x55!−…x3⟹(00)$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{x-{\sin }^{-1}x}{x^3}\\ & =\underset{x\to 0}{lim}\frac{x-(x+\frac{x^3}{3!}+\frac{9x^5}{5!}+\dots )}{x^3}\\ & =\underset{x\to 0}{lim}\frac{-\frac{x^3}{3!}-\frac{9x^5}{5!}-\dots }{x^3}⟹\left(\frac{0}{0}\right)\end{array}$$

Applying L’Hospital rule,

=limx→0−3x26−45x41203x2=limx→0−36−45x21203=−363=−16$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{-\frac{3x^2}{6}-\frac{45x^4}{120}}{3x^2}\\ & =\underset{x\to 0}{lim}\frac{-\frac{3}{6}-\frac{45x^2}{120}}{3}\\ & =\frac{-\frac{3}{6}}{3}\\ & =-\frac{1}{6}\end{array}$$

Alternative method

=limx→0x−sin−1xx3⟹(00)$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{x-{\sin }^{-1}x}{x^3}⟹\left(\frac{0}{0}\right)\end{array}$$

Applying L’Hospital rule,

=limx→01−(1−x2)−1/23x2⟹(00)=limx→012(1−x2)−3/2×(−2x)6x⟹(00)=limx→0−(1−x2)−3/26=−16$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{1-(1-x^2{)}^{-1/2}}{3x^2}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{\frac{1}{2}(1-x^2{)}^{-3/2}\times (-2x)}{6x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{-(1-x^2{)}^{-3/2}}{6}& \\ & =-\frac{1}{6}& \end{array}$$

22. limx→02sinx−sin2xtan3x${lim}_{x\to 0}\frac{2\sin x-\sin 2x}{{\tan }^{3}x}$

This can be written as,

=limx→02sinx−sin2xx3×(tanxx)3=limx→02sinx−sin2xx3×1=limx→02sinx−sin2xx3⟹(00)$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3\times {\left(\frac{\tan x}{x}\right)}^3}\\ & =\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3\times 1}\\ & =\underset{x\to 0}{lim}\frac{2\sin x-\sin 2x}{x^3}⟹\left(\frac{0}{0}\right)\end{array}$$

Applying L’Hospital rule,

=limx→02cosx−2cos2x3x2⟹(00)=limx→0−2sinx+4sin2x6x⟹(00)=limx→0−2cosx+8cos2x6=66=1$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{2\cos x-2\cos 2x}{3x^2}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{-2\sin x+4\sin 2x}{6x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{-2\cos x+8\cos 2x}{6}& \\ & =\frac{6}{6}=1& \end{array}$$

6. limx→11+logx−x1−2x+x2${lim}_{x\to 1}\frac{1+\log x-x}{1-2x+x^2}$

This is also of form 00$\frac{0}{0}$. So applying L’Hospital rule,

=limx→11x−1−2+2x⟹(00)=limx→1−x−22=−12$$\begin{array}{cc}& =\underset{x\to 1}{lim}\frac{\frac{1}{x}-1}{-2+2x}⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 1}{lim}\frac{-x^{-2}}{2}\\ & =-\frac{1}{2}\end{array}$$

7. limx→0log(1−x2)logcosx${lim}_{x\to 0}\frac{\log (1-x^2)}{\log \cos x}$

This is also of form 00$\frac{0}{0}$. So applying L’Hospital rule,

=limx→011−x2×(−2x)−tanx⟹(00)=limx→02(1−x2)+4x2(1−x2)2sec2x=2$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{\frac{1}{1-x^2}\times (-2x)}{-\tan x}⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{\frac{2(1-x^2)+4x^2}{(1-x^2{)}^2}}{{\sec }^{2}x}\\ & =2\end{array}$$

8. limx→0xex−log(1+x)x2${lim}_{x\to 0}\frac{x{e}^x-\log (1+x)}{x^2}$

This is also of form 00$\frac{0}{0}$. So applying L’Hospital rule,

=limx→0xex+ex−11+x2x$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{x{e}^x+e^x-\frac{1}{1+x}}{2x}\end{array}$$

Again of form 00$\frac{0}{0}$, so

=limx→0xex+ex+ex+1x22=0+1+1+12=32$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{x{e}^x+e^x+e^x+\frac{1}{x^2}}{2}\\ & =\frac{0+1+1+1}{2}\\ & =\frac{3}{2}\end{array}$$

9. limx→0tanx−xx2tanx${lim}_{x\to 0}\frac{\tan x-x}{x^2\tan x}$

The expression can be written as,

=limx→0tanx−xx3tanxx=limx→0tanx−xx3, as limx→0tanxx=0=limx→0tanx−xx3⟹(00)=sec2x−13x2⟹(00)=2sec2xtanx6x⟹(00)=sec2xtanx3x⟹(00)=sec4x+2sec2xtan2x3=1+03=13$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{\tan x-x}{x^3\frac{\tan x}{x}}& \\ & =\underset{x\to 0}{lim}\frac{\tan x-x}{x^3},\text{as}\underset{x\to 0}{lim}\frac{\tan x}{x}=0& \\ & =\underset{x\to 0}{lim}\frac{\tan x-x}{x^3}& ⟹\left(\frac{0}{0}\right)\\ & =\frac{{\sec }^{2}x-1}{3x^2}& ⟹\left(\frac{0}{0}\right)\\ & =\frac{2{\sec }^{2}x\tan x}{6x}& ⟹\left(\frac{0}{0}\right)\\ & =\frac{{\sec }^{2}x\tan x}{3x}& ⟹\left(\frac{0}{0}\right)\\ & =\frac{{\sec }^{4}x+2{\sec }^{2}x{\tan }^{2}x}{3}& \\ & =\frac{1+0}{3}=\frac{1}{3}& \end{array}$$

24. limx→0coshx−cosxxsinx${lim}_{x\to 0}\frac{\cos \text{h}x-\cos x}{x\sin x}$

Lets re-write the equation,

=limx→0coshx−cosxx2sinxx=limx→0coshx−cosxx2, as sinxx=1=limx→0coshx−cosxx2⟹(00)$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{\cos \text{h}x-\cos x}{x^2\frac{\sin x}{x}}\\ & =\underset{x\to 0}{lim}\frac{\cos \text{h}x-\cos x}{x^2},\text{as}\frac{\sin x}{x}=1\\ & =\underset{x\to 0}{lim}\frac{\cos \text{h}x-\cos x}{x^2}⟹\left(\frac{0}{0}\right)\end{array}$$

Applying L’Hospital rule,

=sinhx+sinx2x⟹(00)=coshx+cosx2⟹(00)=22=1$$\begin{array}{ccc}& =\frac{\sin \text{h}x+\sin x}{2x}& ⟹\left(\frac{0}{0}\right)\\ & =\frac{\cos \text{h}x+\cos x}{2}& ⟹\left(\frac{0}{0}\right)\\ & =\frac{2}{2}=1& \end{array}$$

11. limx→0(e3x−1)tan2xx3${lim}_{x\to 0}\frac{(e^{3x}-1){\tan }^{2}x}{x^3}$

The expression can be written as,

=limx→0(e3x−1)(tanxx)2×x2x3$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{(e^{3x}-1){\left(\frac{\tan x}{x}\right)}^2\times x^2}{x^3}\end{array}$$

Using property of multiplication of two limits,

=limx→0(e3x−1)x2x3×limx→0(tanxx)2=limx→0(e3x−1)x2x3, as limx→0tanxx=0=limx→0(e3x−1)x⟹(00)$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{(e^{3x}-1)x^2}{x^3}\times \underset{x\to 0}{lim}{\left(\frac{\tan x}{x}\right)}^2& \\ & =\underset{x\to 0}{lim}\frac{(e^{3x}-1)x^2}{x^3},\text{as}\underset{x\to 0}{lim}\frac{\tan x}{x}=0& \\ & =\underset{x\to 0}{lim}\frac{(e^{3x}-1)}{x}& ⟹\left(\frac{0}{0}\right)\end{array}$$

Applying L’Hospital rule,

=limx→03e3x1=3$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{3e^{3x}}{1}\\ & =3\end{array}$$

9.1.2 Question 2

Evaluate the limits:

1. limx→0logsinxcotx${lim}_{x\to 0}\frac{\log \sin x}{\cot x}$

Form of ∞∞$\frac{\infty }{\infty }$, so,

=limx→0cosxsinx−csc2x=limx→0−sinxcosx=0$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{\frac{\cos x}{\sin x}}{-{\csc }^{2}x}\\ & =\underset{x\to 0}{lim}-\sin x\cos x\\ & =0\end{array}$$

2. limx→0logtanxlogx${lim}_{x\to 0}\frac{\log \tan x}{\log x}$

This is a limit of form ∞∞$\frac{\infty }{\infty }$, so

=limx→01tanx×sec2x1x=limx→0xsinxcosx⟹(00)=limx→01cos2x−sin2x=11−0=1$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{\frac{1}{\tan x}\times {\sec }^{2}x}{\frac{1}{x}}& \\ & =\underset{x\to 0}{lim}\frac{x}{\sin x\cos x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{1}{{\cos }^{2}x-{\sin }^{2}x}& \\ & =\frac{1}{1-0}=1& \end{array}$$

3. limx→0logtanxtan2x${lim}_{x\to 0}{\log }_{\tan x}\tan 2x$

Properties of logarithms

• If y=logbx$y={\log }_{b}x$, then x=by$x=b^y$
• logba=logalogb${\log }_{b}a=\frac{\log a}{\log b}$

The given expression can then be written as,

limx→0logtanxtan2x=limx→0logtan2xlogtanx⟹(∞∞)=limx→02sec22xtan2xsec2xtanx=limx→02sinxcosxsin2xcos2x⟹(00)=limx→02(cos2x−sin2x)2(cos22x−sin22x)=22=1$$\begin{array}{ccc}\underset{x\to 0}{lim}{\log }_{\tan x}\tan 2x& =\underset{x\to 0}{lim}\frac{\log \tan 2x}{\log \tan x}& ⟹\left(\frac{\infty }{\infty }\right)\\ & =\underset{x\to 0}{lim}\frac{\frac{2{\sec }^{2}2x}{\tan 2x}}{\frac{{\sec }^{2}x}{\tan x}}& \\ & =\underset{x\to 0}{lim}\frac{2\sin x\cos x}{\sin 2x\cos 2x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{2({\cos }^{2}x-{\sin }^{2}x)}{2({\cos }^{2}2x-{\sin }^{2}2x)}& \\ & =\frac{2}{2}=1& \end{array}$$

4. limx→0xlogx${lim}_{x\to 0}x\log x$

The limit is of form 0×∞$0\times \infty$.

=limx→0logx1x⟹(∞∞)=limx→01x−1x2=limx→0−x=0$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{\log x}{\frac{1}{x}}& ⟹\left(\frac{\infty }{\infty }\right)\\ & =\underset{x\to 0}{lim}\frac{\frac{1}{x}}{-\frac{1}{x^2}}& \\ & =\underset{x\to 0}{lim}-x& \\ & =0& \end{array}$$

22. limx→0xlogtanx${lim}_{x\to 0}x\log \tan x$

This is of form 0×∞$0\times \infty$.

=limx→0logtanx1x⟹(∞∞)=limx→0sec2xtanx−1x2=limx→0−x2sinxcosx⟹(00)=limx→0−2xcos2x−sin2x=01=0$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{\log \tan x}{\frac{1}{x}}& ⟹\left(\frac{\infty }{\infty }\right)\\ & =\underset{x\to 0}{lim}\frac{\frac{{\sec }^{2}x}{\tan x}}{-\frac{1}{x^2}}& \\ & =\underset{x\to 0}{lim}-\frac{x^2}{\sin x\cos x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}-\frac{2x}{{\cos }^{2}x-{\sin }^{2}x}& \\ & =\frac{0}{1}=0& \end{array}$$

6. limx→a(a−x)tan(πx2a)${lim}_{x\to a}(a-x)\tan \left(\frac{\pi x}{2a}\right)$

This is of form 0×∞$0\times \infty$.

=limx→aa−xcotπx2a⟹(00)=limx→a−1−csc2πx2a×π2a=limx→asin2πx2a×2aπ=2aπ$$\begin{array}{ccc}& =\underset{x\to a}{lim}\frac{a-x}{\cot \frac{\pi x}{2a}}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to a}{lim}\frac{-1}{-{\csc }^2\frac{\pi x}{2a}\times \frac{\pi }{2a}}& \\ & =\underset{x\to a}{lim}{\sin }^2\frac{\pi x}{2a}\times \frac{2a}{\pi }& \\ & =\frac{2a}{\pi }& \end{array}$$

7. limx→∞xnex${lim}_{x\to \infty }\frac{x^n}{e^x}$, n$n$ being a positive integer

This is of form ∞∞$\frac{\infty }{\infty }$,

limx→∞xnex=limx→∞nxn−1ex=limx→∞n(n−1)xn−2ex=…=limx→∞n!ex=0$$\begin{array}{cc}\underset{x\to \infty }{lim}\frac{x^n}{e^x}& =\underset{x\to \infty }{lim}\frac{n{x}^{n-1}}{e^x}\\ & =\underset{x\to \infty }{lim}\frac{n(n-1)x^{n-2}}{e^x}=\dots =\underset{x\to \infty }{lim}\frac{n!}{e^x}\\ & =0\end{array}$$

8. limx→0xlogsin2x${lim}_{x\to 0}x\log {\sin }^{2}x$

The expression can be written as, limx→02xlogsinx${lim}_{x\to 0}2x\log \sin x$

This is of form 0×∞$0\times \infty$.

=limx→02logsinx1x⟹(∞∞)=limx→02cotx−1x−2=limx→0−2x2tanx⟹(00)=limx→0−4xsec2x=01=0$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{2\log \sin x}{\frac{1}{x}}& ⟹\left(\frac{\infty }{\infty }\right)\\ & =\underset{x\to 0}{lim}\frac{2\cot x}{-1x^{-2}}& \\ & =\underset{x\to 0}{lim}-\frac{2x^2}{\tan x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}-\frac{4x}{{\sec }^{2}x}& \\ & =\frac{0}{1}=0& \end{array}$$

9.1.3 Question 3

Evaluate the limits:

1. limx→0(1x2−1sin2x)${lim}_{x\to 0}(\frac{1}{x^2}-\frac{1}{{\sin }^{2}x})$ [TU 2063, 64]

This is of form ∞−∞$\infty -\infty$.

=limx→0sin2x−x2x2sin2x=limx→0sin2x−x2x4×(sinxx)2=limx→0sin2x−x2x4,limx→0(sinxx)=1=limx→0sin2x−x2x4⟹(00)=limx→02sinxcosx−2x4x3⟹(00)=limx→02(cos2x−sin2x)−212x2⟹(00)=limx→02cos2x−212x2⟹(00)=limx→0−4sin2x24x⟹(00)=limx→0−8cos2x24=−13$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{{\sin }^{2}x-x^2}{x^2{\sin }^{2}x}& \\ & =\underset{x\to 0}{lim}\frac{{\sin }^{2}x-x^2}{x^4\times {\left(\frac{\sin x}{x}\right)}^2}& \\ & =\underset{x\to 0}{lim}\frac{{\sin }^{2}x-x^2}{x^4},\underset{x\to 0}{lim}\left(\frac{\sin x}{x}\right)=1& \\ & =\underset{x\to 0}{lim}\frac{{\sin }^{2}x-x^2}{x^4}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{2\sin x\cos x-2x}{4x^3}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{2({\cos }^{2}x-{\sin }^{2}x)-2}{12x^2}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{2\cos 2x-2}{12x^2}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{-4\sin 2x}{24x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{-8\cos 2x}{24}& \\ & =-\frac{1}{3}& \end{array}$$

2. limx→0(1x−1ex−1)${lim}_{x\to 0}(\frac{1}{x}-\frac{1}{e^x-1})$

This is of form ∞−∞$\infty -\infty$.

=limx→0ex−x−1x(ex−1)=limx→0ex−x−1x2×ex−1x=limx→0ex−x−1x2⟹limx→0ex−1x=1=limx→0ex−x−1x2⟹(00)=limx→0ex−12x⟹(00)=limx→0ex2=12$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{e^x-x-1}{x(e^x-1)}& \\ & =\underset{x\to 0}{lim}\frac{e^x-x-1}{x^2\times \frac{e^x-1}{x}}& \\ & =\underset{x\to 0}{lim}\frac{e^x-x-1}{x^2}& ⟹\underset{x\to 0}{lim}\frac{e^x-1}{x}=1\\ & =\underset{x\to 0}{lim}\frac{e^x-x-1}{x^2}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{e^x-1}{2x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{e^x}{2}& \\ & =\frac{1}{2}& \end{array}$$

3. limx→0(1x2−cot2x)${lim}_{x\to 0}(\frac{1}{x^2}-{\cot }^{2}x)$

This is of form ∞−∞$\infty -\infty$.

=limx→01x2−cos2xsin2x=sin2x−x2cos2xx2sin2x$$\begin{array}{cc}& =\underset{x\to 0}{lim}\frac{1}{x^2}-\frac{{\cos }^{2}x}{{\sin }^{2}x}\\ & =\frac{{\sin }^{2}x-x^2{\cos }^{2}x}{x^2{\sin }^{2}x}\end{array}$$

After we put the expansion of sinx$\sin x$ and cosx$\cos x$, we get,

=(x−x33!+x55!−x77!+…)2−x2(1−x22!+x44!−x66!+…)2x2(x−x33!+x55!−x77!+…)2=limx→023x4+…x4+…=23$$\begin{array}{cc}& =\frac{{(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots )}^2-x^2{(1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\dots )}^2}{x^2{(x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\dots )}^2}\\ & =\underset{x\to 0}{lim}\frac{\frac{2}{3}{x}^4+\dots }{x^4+\dots }\\ & =\frac{2}{3}\end{array}$$

4. limx→0(1x−1x2log(1+x))${lim}_{x\to 0}(\frac{1}{x}-\frac{1}{x^2}\log (1+x\left)\right)$

This is of form ∞−∞$\infty -\infty$.

=limx→0x−log(1+x)x2⟹(00)=limx→01−11+x2x⟹(00)=limx→0(1+x)−22=12$$\begin{array}{ccc}& =\underset{x\to 0}{lim}\frac{x-\log (1+x)}{x^2}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{1-\frac{1}{1+x}}{2x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{(1+x{)}^{-2}}{2}& \\ & =\frac{1}{2}& \end{array}$$

9.1.4 Question 4

Evaluate the limits:

1. limx→0xx${lim}_{x\to 0}{x}^x$

This is of form 00$0^0$.

y=xxlogy=logxxlimx→0logy=limx→0xlogx⟹0×∞=limx→0logx1x⟹(∞∞)$$\begin{array}{ccc}y& =x^x& \\ \log y& =\log x^x& \\ \underset{x\to 0}{lim}\log y& =\underset{x\to 0}{lim}x\log x& ⟹0\times \infty \\ & =\underset{x\to 0}{lim}\frac{\log x}{\frac{1}{x}}& ⟹\left(\frac{\infty }{\infty }\right)\end{array}$$

Applying L’Hospital rule,

limx→0logy=limx→01x−x−2=limx→0−xlimx→0logy=0log(limx→0y)=0limx→0y=e0limx→0xx=1$$\begin{array}{cc}\underset{x\to 0}{lim}\log y& =\underset{x\to 0}{lim}\frac{\frac{1}{x}}{-x^{-2}}\\ & =\underset{x\to 0}{lim}-x\\ \underset{x\to 0}{lim}\log y& =0\\ \log \left(\underset{x\to 0}{lim}y\right)& =0\\ \underset{x\to 0}{lim}y& =e^0\\ \underset{x\to 0}{lim}{x}^x& =1\end{array}$$

2. limx→π/2(sinx)tanx${lim}_{x\to \pi /2}(\sin x{)}^{\tan x}$

limx→0logy=log(limx→0y)$$\begin{array}{cc}\underset{x\to 0}{lim}\log y& =\log \left(\underset{x\to 0}{lim}y\right)\end{array}$$

The limit is of form 1∞$1^{\infty }$.

y=(sinx)tanxlogy=log(sinx)tanxlogy=tanxlogsinxlimx→π/2logy=limx→π/2logsinxcotx⟹(00)=limx→π/2cosxsinx−csc2x=limx→π/2−sinxcosxlimx→π/2logy=0log(limx→π/2y)=0limx→π/2y=e0limx→π/2(sinx)tanx=1$$\begin{array}{ccc}y& =(\sin x{)}^{\tan x}& \\ \log y& =\log (\sin x{)}^{\tan x}& \\ \log y& =\tan x\log \sin x& \\ \underset{x\to \pi /2}{lim}\log y& =\underset{x\to \pi /2}{lim}\frac{\log \sin x}{\cot x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to \pi /2}{lim}\frac{\frac{\cos x}{\sin x}}{-{\csc }^{2}x}& \\ & =\underset{x\to \pi /2}{lim}-\sin x\cos x& \\ \underset{x\to \pi /2}{lim}\log y& =0& \\ \log \left(\underset{x\to \pi /2}{lim}y\right)& =0& \\ \underset{x\to \pi /2}{lim}y& =e^0& \\ \underset{x\to \pi /2}{lim}(\sin x{)}^{\tan x}& =1& \end{array}$$

3. limx→0(cotx)sin2x${lim}_{x\to 0}(\cot x{)}^{\sin 2x}$

It is of form ∞0${\infty }^0$.

y=(cotx)sin2xlogy=sin2xlogcotxlimx→0logy=limx→0sin2xlogcotx⟹0×∞=limx→0logcot2xcsc2x⟹(∞∞)=limx→0−2csc22xcot2x−2csc2xcot2x=limx→0csc2xcot2x×1cot2x=limx→0tan2xcos2x=01limx→0logy=0limx→0y=e0limx→0(cotx)sin2x=1$$\begin{array}{ccc}y& =(\cot x{)}^{\sin 2x}& \\ \log y& =\sin 2x\log \cot x& \\ \underset{x\to 0}{lim}\log y& =\underset{x\to 0}{lim}\sin 2x\log \cot x& ⟹0\times \infty \\ & =\underset{x\to 0}{lim}\frac{\log \cot 2x}{\csc 2x}& ⟹\left(\frac{\infty }{\infty }\right)\\ & =\underset{x\to 0}{lim}\frac{\frac{-2{\csc }^{2}2x}{\cot 2x}}{-2\csc 2x\cot 2x}& \\ & =\underset{x\to 0}{lim}\frac{\csc 2x}{\cot 2x}\times \frac{1}{\cot 2x}& \\ & =\underset{x\to 0}{lim}\frac{\tan 2x}{\cos 2x}=\frac{0}{1}& \\ \underset{x\to 0}{lim}\log y& =0& \\ \underset{x\to 0}{lim}y& =e^0& \\ \underset{x\to 0}{lim}(\cot x{)}^{\sin 2x}& =1& \end{array}$$

4. limx→0(cosx)cot2x${lim}_{x\to 0}(\cos x{)}^{{\cot }^{2}x}$ [TU 2058]

It is of form 1∞$1^{\infty }$.

y=(cosx)cot2xlogy=cot2xlogcosxlimx→0logy=limx→0cot2xlogcosx⟹∞×0=limx→0logcosxtan2x=limx→0logcosxx2(tanxx)2=limx→0logcosxx2⟹(00)=limx→0−tanx2x⟹(00)=limx→0−sec2x2limx→0logy=−12limx→0(cosx)cot2x=e−1/2$$\begin{array}{ccc}y& =(\cos x{)}^{{\cot }^{2}x}& \\ \log y& ={\cot }^{2}x\log \cos x& \\ \underset{x\to 0}{lim}\log y& =\underset{x\to 0}{lim}{\cot }^{2}x\log \cos x& ⟹\infty \times 0\\ & =\underset{x\to 0}{lim}\frac{\log \cos x}{{\tan }^{2}x}& \\ & =\underset{x\to 0}{lim}\frac{\log \cos x}{x^2{\left(\frac{\tan x}{x}\right)}^2}& \\ & =\underset{x\to 0}{lim}\frac{\log \cos x}{x^2}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{-\tan x}{2x}& ⟹\left(\frac{0}{0}\right)\\ & =\underset{x\to 0}{lim}\frac{-{\sec }^{2}x}{2}& \\ \underset{x\to 0}{lim}\log y& =-\frac{1}{2}& \\ \underset{x\to 0}{lim}(\cos x{)}^{{\cot }^{2}x}& =e^{-1/2}& \end{array}$$

22. limx→π(sinx)tanx

It is of form 0∞.

y=(sinx)tanxlogy=tanxlogsinxlimx→πlogy=limx→πtanxlogsinx⟹0×∞=limx→πlogsinxcotx⟹(∞∞)=limx→πcotx−csc2x=limx→π−sinxcosxlimx→πlogy=0limx→πy=e0limx→π(sinx)tanx=1

6. limx→0(cot2x)sinx [TU 2054, 2060, 2066]

The limit is of form ∞0.

y=(cot2x)sinxlogy=sinxlog(cot2x)limx→0logy=limx→0sinxlog(cot2x)⟹0×∞=limx→0log(cot2x)cscx⟹(∞∞)=limx→01cot2x×2cotx×(−csc2x)−cscxcotx=limx→02sinxcos2xlimx→0logy=0limx→0y=e0limx→0(cot2x)sinx=1

7. limx→0(1x2)tanx

The limit will be of form ∞0.

y=(1x2)tanxlogy=tanxlog(1x2)=tanx(log1−logx2)=−2logxtanxlimx→0logy=limx→0−2logxtanx⟹∞×0=limx→0−2logxcotx⟹(∞∞)=limx→0−2x−csc2x=limx→02sin2xx=limx→02x(sinxx)2=limx→02xlimx→0logy=0limx→0y=e0limx→0(1x2)tanx=1

8. limx→0(tanxx)1/x [TU 2057, 2062, 2063]

Let,

y=(tanxx)1/xlogy=1xlog(tanxx)limx→0logy=limx→01xlog(tanxx)=limx→0log(tanxx)x⟹(00)=limx→01tanxx×ddx(tanxx)1⟹limx→0tanxx=1=limx→011×ddx(tanxx)=limx→0xsec2x−tanxx2⟹(00)=limx→02xsec2xtanx+sec2x−sec2x2x=limx→0sec2xtanxlimx→0logy=0limx→0y=e0limx→0(tanxx)1/x=1

9. limx→0(sinxx)1/x2

This is of form 1∞.

y=(sinxx)1/x2logy=1x2log(sinxx)limx→0logy=limx→0log(sinxx)x2⟹(00)=limx→01sinxx×ddx(sinxx)2x=limx→01×xcosx−sinxx22x=limx→0xcosx−sinx2x3⟹(00)=limx→0−xsinx+cosx−cosx6x2=limx→0−sinx6x⟹(00)=limx→0−cosx6limx→0logy=−16limx→0y=e−1/6limx→0(sinxx)1/x2=e−1/6

24. limx→0(cotx)1/logx

This limit is of form ∞0.

y=(cotx)1/logxlogy=1logxlog(cotx)limx→0logy=limx→0log(cotx)logx⟹(∞∞)=limx→01cotx×(−csc2x)1x=limx→0−xsinxcosx⟹(00)=limx→0−1cos2x−sin2x=−1limx→0logy=−1limx→0y=e−1limx→0(cotx)1/logx=e−1

11. limx→0(tanxx)1/x2

The limit of this expression is 1∞.

y=(tanxx)1/x2logy=1x2log(tanxx)limx→0logy=limx→0log(tanxx)x2⟹(00)=limx→01tanxx×ddx(tanxx)2x⟹limx→0tanxx=1=limx→0xsec2x−tanx2x×x2⟹(00)=limx→02xsec2xtanx+sec2x−sec2x6x2=limx→0sec2xtanx3x⟹limx→0sec2x=1=limx→01×tanx3x⟹(00)=limx→0sec2x3limx→0logy=13limx→0y=e1/3limx→0(tanxx)1/x2=e1/3