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NCERT Exemplar · Class 11 Mathematics Limits and Derivatives

80 questions · 80 still being checked

EXERCISE 13.3 11–20 (part 2 of 8)

  1. Evaluate :

    Exercise 11

    lim⁡x→01+x3−1−x3x2\displaystyle \lim _{x \rightarrow 0} \frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2}

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    NCERT’s answer
    $\displaystyle 0$
    Rationalise the numerator.\[\frac{\sqrt{1+x^3}-\sqrt{1-x^3}}{x^2}\cdot\frac{\sqrt{1+x^3}+\sqrt{1-x^3}}{\sqrt{1+x^3}+\sqrt{1-x^3}}=\frac{(1+x^3)-(1-x^3)}{x^2\left(\sqrt{1+x^3}+\sqrt{1-x^3}\right)} \]\[=\frac{2x}{\sqrt{1+x^3}+\sqrt{1-x^3}} \]\[\lim_{x\to0}\frac{2x}{\sqrt{1+x^3}+\sqrt{1-x^3}}=\frac{0}{1+1}=0 \]Answer: \(\displaystyle 0\)

  2. Exercise 12

    lim⁡x→−3x3+27x5+243\displaystyle \lim _{x \rightarrow-3} \frac{x^3+27}{x^5+243}

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    NCERT’s answer
    \(\displaystyle \frac{1}{15}\)
    Write both terms as \(\displaystyle x^n-a^n\) with \(\displaystyle a=-3\), and use \(\displaystyle \lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}\).\[\frac{x^3+27}{x^5+243}=\frac{x^3-(-3)^3}{x^5-(-3)^5}=\frac{\dfrac{x^3-(-3)^3}{x+3}}{\dfrac{x^5-(-3)^5}{x+3}} \]\[\lim_{x\to-3}\frac{x^3+27}{x^5+243}=\frac{3(-3)^2}{5(-3)^4}=\frac{27}{405}=\frac{1}{15} \]Answer: \(\displaystyle \dfrac{1}{15}\)

  3. Exercise 13

    lim⁡x→12(8x−32x−1−4x2+14x2−1)\displaystyle \lim _{x \rightarrow \frac{1}{2}}\left(\frac{8 x-3}{2 x-1}-\frac{4 x^2+1}{4 x^2-1}\right)

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    NCERT’s answer
    \(\displaystyle \frac{7}{2}\)
    Combine over the common denominator \(\displaystyle 4x^2-1=(2x-1)(2x+1)\).\[\frac{8x-3}{2x-1}-\frac{4x^2+1}{4x^2-1}=\frac{(8x-3)(2x+1)-(4x^2+1)}{(2x-1)(2x+1)} \]\[=\frac{12x^2+2x-4}{(2x-1)(2x+1)}=\frac{2(2x-1)(3x+2)}{(2x-1)(2x+1)}=\frac{2(3x+2)}{2x+1} \]\[\lim_{x\to\frac12}\frac{2(3x+2)}{2x+1}=\frac{2\left(\frac32+2\right)}{1+1}=\frac{7}{2} \]Answer: \(\displaystyle \dfrac{7}{2}\)

  4. Exercise 14

    Find ‘n\displaystyle n’, if lim⁡x→2xn−2nx−2=80,n∈N\displaystyle \lim _{x \rightarrow 2} \frac{x^n-2^n}{x-2}=80, n \in \mathbf{N}

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    NCERT’s answer
    \(\displaystyle n = 5\)
    Use \(\displaystyle \lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}\) with \(\displaystyle a=2\).\[n\cdot2^{\,n-1}=80 \]The left side increases with \(\displaystyle n\), so test consecutive values.\[4\cdot2^{3}=32,\qquad 5\cdot2^{4}=80,\qquad 6\cdot2^{5}=192 \]Answer: \(\displaystyle n=5\)
  5. Evaluate :

    Exercise 15

    lim⁡x→asin⁡3xsin⁡7x\displaystyle \lim _{x \rightarrow a} \frac{\sin 3 x}{\sin 7 x}

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    Case \(\displaystyle \sin 7a\neq0\). Both sines are continuous at \(\displaystyle a\); substitute.\[\lim_{x\to a}\frac{\sin3x}{\sin7x}=\frac{\sin3a}{\sin7a} \]Case \(\displaystyle a=n\pi\) (including \(\displaystyle a=0\)). Put \(\displaystyle x=a+t\); both sines pick up the same sign \(\displaystyle (-1)^n\).\[\lim_{t\to0}\frac{\sin3t}{\sin7t}=\lim_{t\to0}\frac{3}{7}\cdot\frac{\sin3t}{3t}\cdot\frac{7t}{\sin7t}=\frac37 \]Case \(\displaystyle a=\dfrac{k\pi}{7}\), \(\displaystyle 7\nmid k\). Here \(\displaystyle \sin7x\to0\) but \(\displaystyle \sin3a\neq0\), so the quotient is unbounded and there is no limit.Answer: \(\displaystyle \dfrac{\sin3a}{\sin7a}\) if \(\displaystyle \sin7a\neq0\); \(\displaystyle \dfrac37\) if \(\displaystyle a=n\pi\) (in particular \(\displaystyle a=0\)); no limit otherwise.

  6. Exercise 16

    lim⁡x→0sin⁡22xsin⁡24x\displaystyle \lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{\sin ^2 4 x}

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    NCERT’s answer
    \(\displaystyle \frac{1}{4}\)
    Use the double-angle identity on the denominator.\[\sin4x=2\sin2x\cos2x \]\[\frac{\sin^22x}{\sin^24x}=\frac{\sin^22x}{4\sin^22x\cos^22x}=\frac{1}{4\cos^22x} \]\[\lim_{x\to0}\frac{1}{4\cos^22x}=\frac{1}{4\cdot1}=\frac14 \]Answer: \(\displaystyle \dfrac14\)

  7. Exercise 17

    lim⁡x→01−cos⁡2xx2\displaystyle \lim _{x \rightarrow 0} \frac{1-\cos 2 x}{x^2}

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    NCERT’s answer
    $\displaystyle 2$
    \[1-\cos 2x = 2\sin^2 x \] \[\lim_{x\to0}\frac{1-\cos 2x}{x^2} = 2\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2 \] \[= 2(1)^2 = 2 \] Answer: \(\displaystyle 2\)

  8. Exercise 18

    lim⁡x→02sin⁡x−sin⁡2xx3\displaystyle \lim _{x \rightarrow 0} \frac{2 \sin x-\sin 2 x}{x^3}

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    NCERT’s answer
    $\displaystyle 1$
    \[2\sin x-\sin 2x = 2\sin x-2\sin x\cos x = 2\sin x\,(1-\cos x) \] \[\frac{2\sin x-\sin 2x}{x^3} = 2\cdot\frac{\sin x}{x}\cdot\frac{1-\cos x}{x^2} \] \[\frac{1-\cos x}{x^2} = \frac{2\sin^2\dfrac{x}{2}}{x^2} = \frac12\left(\frac{\sin\dfrac{x}{2}}{\dfrac{x}{2}}\right)^2 \to \frac12 \] \[\lim_{x\to0}\frac{2\sin x-\sin 2x}{x^3} = 2\cdot 1\cdot\frac12 = 1 \] Answer: \(\displaystyle 1\)

  9. Exercise 19

    lim⁡x→01−cos⁡mx1−cos⁡nx\displaystyle \lim _{x \rightarrow 0} \frac{1-\cos m x}{1-\cos n x}

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    NCERT’s answer
    \(\displaystyle \frac{m^2}{n^2}\)
    \[1-\cos mx = 2\sin^2\frac{mx}{2}, \qquad 1-\cos nx = 2\sin^2\frac{nx}{2} \] \[\frac{1-\cos mx}{1-\cos nx} = \frac{\left(\dfrac{\sin\dfrac{mx}{2}}{\dfrac{mx}{2}}\right)^2\cdot\dfrac{m^2x^2}{4}}{\left(\dfrac{\sin\dfrac{nx}{2}}{\dfrac{nx}{2}}\right)^2\cdot\dfrac{n^2x^2}{4}} \] \[\lim_{x\to0}\frac{1-\cos mx}{1-\cos nx} = \frac{1\cdot m^2}{1\cdot n^2} = \frac{m^2}{n^2} \quad (n\neq 0) \] Answer: \(\displaystyle \dfrac{m^2}{n^2}\)

  10. Exercise 20

    lim⁡x→π31−cos⁡6x2(π3−x)\displaystyle \lim _{x \rightarrow \frac{\pi}{3}} \frac{\sqrt{1-\cos 6 x}}{\sqrt{2}\left(\dfrac{\pi}{3}-x\right)}

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    NCERT’s printed answer for this exercise does not match its own question. This working follows the question as printed.

    \[1-\cos 6x = 2\sin^2 3x \;\Rightarrow\; \sqrt{1-\cos 6x} = \sqrt2\,|\sin 3x| \] \[\frac{\sqrt{1-\cos 6x}}{\sqrt2\left(\dfrac{\pi}{3}-x\right)} = \frac{|\sin 3x|}{\dfrac{\pi}{3}-x} \] Put \(\displaystyle h=\frac{\pi}{3}-x\), so \(\displaystyle \sin 3x=\sin(\pi-3h)=\sin 3h\): \[\frac{|\sin 3h|}{h} \] \[x\to\tfrac{\pi}{3}^-\ (h\to0^+):\quad \frac{\sin 3h}{h} = 3\cdot\frac{\sin 3h}{3h}\to 3 \] \[x\to\tfrac{\pi}{3}^+\ (h\to0^-):\quad -\frac{\sin 3h}{h}\to -3 \] The one-sided limits differ, so the limit does not exist. The value \(\displaystyle 3\) results only if \(\displaystyle |\sin 3x|\) is read as \(\displaystyle \sin 3x\), which holds for \(\displaystyle x<\frac{\pi}{3}\).Answer: the limit does not exist; the left-hand limit is \(\displaystyle 3\) and the right-hand limit is \(\displaystyle -3\).