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NCERT Exemplar · Class 11 Mathematics Limits and Derivatives

80 questions · 80 still being checked

EXERCISE 13.3 51–60 (part 6 of 8)

  1. Evaluate each of the following limits in Exercises $\displaystyle 47$ to 53.

    Exercise 51

    Show that lim⁡x→4∣x−4∣x−4\displaystyle \lim _{x \rightarrow 4} \frac{|x-4|}{x-4} does not exists

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    The limit exists only if the two one-sided limits are equal. \[x>4:\quad |x-4|=x-4 \ \Rightarrow\ \frac{|x-4|}{x-4}=1 \] \[\lim_{x\to4^+}\frac{|x-4|}{x-4}=1 \] \[x<4:\quad |x-4|=-(x-4) \ \Rightarrow\ \frac{|x-4|}{x-4}=-1 \] \[\lim_{x\to4^-}\frac{|x-4|}{x-4}=-1 \] \[\lim_{x\to4^-}\neq\lim_{x\to4^+} \] Answer: The limit does not exist (left limit \(\displaystyle -1\), right limit \(\displaystyle 1\)).

  2. Exercise 52

    Let f(x)={kcos⁡xπ−2x when x≠π23x=π2\displaystyle f(x)=\left\{\begin{array}{cc}\frac{k \cos x}{\pi-2 x} & \text { when } x \neq \frac{\pi}{2} \\ 3 & x=\frac{\pi}{2}\end{array}\right. and if lim⁡x→π2f(x)=f(π2)\displaystyle \lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right), find the value of k\displaystyle k.

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    NCERT’s answer
    \(\displaystyle k=6\)
    Put \(\displaystyle t=\frac{\pi}{2}-x\); then \(\displaystyle t\to0\) as \(\displaystyle x\to\frac{\pi}{2}\). \[\cos x=\sin t,\qquad \pi-2x=2t \] \[\lim_{x\to\frac{\pi}{2}}f(x)=\lim_{t\to0}\frac{k\sin t}{2t}=\frac{k}{2} \] \[\frac{k}{2}=f\left(\frac{\pi}{2}\right)=3 \] \[k=6 \] Answer: \(\displaystyle k=6\)

  3. Exercise 53

    Let f(x)={x+2x≤1cx2x>−1\displaystyle f(x)=\left\{\begin{array}{cc}x+2 & x \leq 1 \\ c x^2 & x>-1\end{array}\right., find ‘c\displaystyle c’ if lim⁡x→−1f(x)\displaystyle \lim _{x \rightarrow-1} f(x) exists.

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    NCERT’s answer
    \(\displaystyle c=1\)
    The printed \(\displaystyle x\le1\) overlaps \(\displaystyle x>-1\); read it as \(\displaystyle x\le-1\). \[\lim_{x\to-1^-}f(x)=\lim_{x\to-1^-}(x+2)=1 \] \[\lim_{x\to-1^+}f(x)=\lim_{x\to-1^+}cx^2=c \] \[\lim_{x\to-1^-}f(x)=\lim_{x\to-1^+}f(x)\ \Rightarrow\ c=1 \] Answer: \(\displaystyle c=1\)
  4. Choose the correct answer out of $\displaystyle 4$ options given against each Exercise $\displaystyle 54$ to $\displaystyle 76$ (M.C.Q).

    Exercise 54

    lim⁡x→πsin⁡xx−π\displaystyle \lim _{x \rightarrow \pi} \frac{\sin x}{x-\pi} is
    (A)
    1\displaystyle 1 (B) 2\displaystyle 2 (C) -1\displaystyle 1 (D) -2\displaystyle 2

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    NCERT’s answer
    C
    (C) \(\displaystyle -1\). Put \(\displaystyle h=x-\pi\) and use \(\displaystyle \sin(\pi+h)=-\sin h\). \[\lim_{x\to\pi}\frac{\sin x}{x-\pi}=\lim_{h\to0}\frac{-\sin h}{h}=-1 \]

  5. Exercise 55

    lim⁡x→0x2cos⁡x1−cos⁡x\displaystyle \lim _{x \rightarrow 0} \frac{x^2 \cos x}{1-\cos x} is
    (A)
    2\displaystyle 2 (B) 32\displaystyle \frac{3}{2}
    (C)
    −32\displaystyle \frac{-3}{2}

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    NCERT’s answer
    A
    (A) \(\displaystyle 2\). Use \(\displaystyle 1-\cos x=2\sin^2\frac{x}{2}\). \[\frac{x^2\cos x}{1-\cos x}=\frac{x^2\cos x}{2\sin^2\dfrac{x}{2}}=2\cos x\left(\frac{x/2}{\sin\dfrac{x}{2}}\right)^2 \] \[\lim_{x\to0}=2\cdot1\cdot1^2=2 \]

  6. Exercise 56

    lim⁡x→0(1+x)n−1x\displaystyle \lim _{x \rightarrow 0} \frac{(1+x)^n-1}{x} is
    (A)
    n\displaystyle n (B) 1\displaystyle 1 (C) −n\displaystyle -n (D) 0\displaystyle 0

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    NCERT’s answer
    A
    (A) \(\displaystyle n\). Put \(\displaystyle y=1+x\); then \(\displaystyle y\to1\). \[\lim_{x\to0}\frac{(1+x)^n-1}{x}=\lim_{y\to1}\frac{y^n-1^n}{y-1} \] \[\lim_{y\to a}\frac{y^n-a^n}{y-a}=na^{n-1}\ \Rightarrow\ n\cdot1^{n-1}=n \]

  7. Exercise 57

    lim⁡x→1xm−1xn−1\displaystyle \lim _{x \rightarrow 1} \frac{x^m-1}{x^n-1} is
    (A)
    1\displaystyle 1 (B) mn\displaystyle \frac{m}{n}
    (C)
    −mn\displaystyle -\frac{m}{n}
    (D)
    m2n2\displaystyle \frac{m^2}{n^2}

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    NCERT’s answer
    B
    (B) \(\displaystyle \dfrac{m}{n}\) Divide numerator and denominator by \(\displaystyle x-1\). \[\frac{x^m-1}{x^n-1}=\frac{(x^m-1)/(x-1)}{(x^n-1)/(x-1)} \] \[\lim_{x\to1}\frac{x^k-1}{x-1}=k\cdot 1^{k-1}=k \] \[\lim_{x\to1}\frac{x^m-1}{x^n-1}=\frac{m}{n} \]

  8. Exercise 58

    lim⁡x→01−cos⁡4θ1−cos⁡6θ\displaystyle \lim _{x \rightarrow 0} \frac{1-\cos 4 \theta}{1-\cos 6 \theta} is
    (A)
    49\displaystyle \frac{4}{9}
    (B)
    12\displaystyle \frac{1}{2}
    (C)
    −12\displaystyle \frac{-1}{2}

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    NCERT’s answer
    A
    (A) \(\displaystyle \dfrac{4}{9}\) Take \(\displaystyle \theta\to0\) and use \(\displaystyle 1-\cos 2A=2\sin^2A\). \[\frac{1-\cos4\theta}{1-\cos6\theta}=\frac{2\sin^2 2\theta}{2\sin^2 3\theta} \] \[=\frac{4}{9}\cdot\frac{\left(\dfrac{\sin 2\theta}{2\theta}\right)^2}{\left(\dfrac{\sin 3\theta}{3\theta}\right)^2} \] \[\to\frac{4}{9}\cdot\frac{1^2}{1^2}=\frac{4}{9} \]

  9. Exercise 59

    lim⁡x→0cosec⁡x−cot⁡xx\displaystyle \lim _{x \rightarrow 0} \frac{\operatorname{cosec} x-\cot x}{x} is
    (A)
    −12\displaystyle \frac{-1}{2}
    (B)
    1\displaystyle 1 (C) 12\displaystyle \frac{1}{2}

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    NCERT’s answer
    C
    (C) \(\displaystyle \dfrac12\) \[\operatorname{cosec}x-\cot x=\frac{1-\cos x}{\sin x} \] \[\frac{1-\cos x}{x\sin x}=\frac{2\sin^2\frac x2}{x\cdot2\sin\frac x2\cos\frac x2}=\frac{\tan\frac x2}{x} \] \[=\frac12\cdot\frac{\tan\frac x2}{\frac x2}\to\frac12\cdot1=\frac12 \]

  10. Exercise 60

    lim⁡x→0sin⁡xx+1−1−x\displaystyle \lim _{x \rightarrow 0} \frac{\sin x}{\sqrt{x+1}-\sqrt{1-x}} is
    (A)
    2\displaystyle 2 (B) 0\displaystyle 0 (C) 1\displaystyle 1 (D) -1\displaystyle 1

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    NCERT’s answer
    C
    (C) \(\displaystyle 1\) Rationalise the denominator. \[\frac{\sin x}{\sqrt{1+x}-\sqrt{1-x}}=\frac{\sin x\left(\sqrt{1+x}+\sqrt{1-x}\right)}{(1+x)-(1-x)} \] \[=\frac{\sin x}{x}\cdot\frac{\sqrt{1+x}+\sqrt{1-x}}{2} \] \[\to 1\cdot\frac{1+1}{2}=1 \]