SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Limits and Derivatives

73 questions · 73 still being checked

EXERCISE 12.1 11–20 (part 2 of 7)

  1. Evaluate the following limits in Exercises $\displaystyle 1$ to 22.

    Exercise 11

    lim⁡x→1ax2+bx+ccx2+bx+a,a+b+c≠0\displaystyle \lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}, a+b+c \neq 0

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    NCERT’s answer
    $\displaystyle 1$
    Direct substitution. At \(\displaystyle x = 1\) the denominator becomes \(\displaystyle c+b+a = a+b+c\), which is given to be non-zero, so the quotient rule applies.\[\lim_{x \to 1} \frac{ax^{2}+bx+c}{cx^{2}+bx+a} = \frac{a+b+c}{c+b+a} = 1 \]The limit is \(\displaystyle 1\).

  2. Exercise 12

    lim⁡x→−21x+12x+2\displaystyle \lim _{x \rightarrow-2} \frac{\dfrac{1}{x}+\dfrac{1}{2}}{x+2}

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    NCERT’s answer
    $\displaystyle -\frac{1}{4}$
    Simplify the compound fraction, then cancel. The form is \(\displaystyle \frac{0}{0}\), so combine the two terms in the numerator first.\[\frac{1}{x}+\frac{1}{2} = \frac{2+x}{2x} \]Hence, for \(\displaystyle x \neq -2\),\[\frac{\dfrac{1}{x}+\dfrac{1}{2}}{x+2} = \frac{x+2}{2x(x+2)} = \frac{1}{2x} \]\[\lim_{x \to -2} \frac{\dfrac{1}{x}+\dfrac{1}{2}}{x+2} = \lim_{x \to -2} \frac{1}{2x} = \frac{1}{-4} = -\frac{1}{4} \]The limit is \(\displaystyle -\dfrac{1}{4}\).

  3. Exercise 13

    lim⁡x→0sin⁡axbx\displaystyle \lim _{x \rightarrow 0} \frac{\sin a x}{b x}

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    NCERT’s answer
    $\displaystyle \frac{a}{b}$
    Standard limit \(\displaystyle \lim\limits_{\theta \to 0}\dfrac{\sin \theta}{\theta} = 1\).Force the angle \(\displaystyle ax\) to appear under the sine as well as below it:\[\frac{\sin ax}{bx} = \frac{a}{b}\cdot\frac{\sin ax}{ax} \qquad (a \neq 0) \]As \(\displaystyle x \to 0\), \(\displaystyle ax \to 0\), so \(\displaystyle \dfrac{\sin ax}{ax} \to 1\) and\[\lim_{x \to 0} \frac{\sin ax}{bx} = \frac{a}{b} \]The limit is \(\displaystyle \dfrac{a}{b}\). (If \(\displaystyle a = 0\) the expression is identically \(\displaystyle 0\), which \(\displaystyle \frac{a}{b}\) also gives.)

  4. Exercise 14

    lim⁡x→0sin⁡axsin⁡bx,a,b≠0\displaystyle \lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}, a, b \neq 0

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    NCERT’s answer
    $\displaystyle \frac{a}{b}$
    Standard limit \(\displaystyle \lim\limits_{\theta \to 0}\dfrac{\sin \theta}{\theta} = 1\), applied twice.Write the quotient so that each sine sits over its own angle:\[\frac{\sin ax}{\sin bx} = \frac{a}{b}\cdot\frac{\dfrac{\sin ax}{ax}}{\dfrac{\sin bx}{bx}} \]As \(\displaystyle x \to 0\), both \(\displaystyle ax \to 0\) and \(\displaystyle bx \to 0\), so each bracketed quotient tends to \(\displaystyle 1\):\[\lim_{x \to 0} \frac{\sin ax}{\sin bx} = \frac{a}{b}\cdot\frac{1}{1} = \frac{a}{b} \]The limit is \(\displaystyle \dfrac{a}{b}\).

  5. Exercise 15

    lim⁡x→πsin⁡(π−x)π(π−x)\displaystyle \lim _{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)}

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    NCERT’s answer
    $\displaystyle \frac{1}{\pi}$
    Change of variable to reach \(\displaystyle \lim\limits_{y \to 0}\dfrac{\sin y}{y} = 1\).Put \(\displaystyle y = \pi - x\). As \(\displaystyle x \to \pi\), \(\displaystyle y \to 0\), and the expression becomes\[\lim_{x \to \pi} \frac{\sin(\pi-x)}{\pi(\pi-x)} = \lim_{y \to 0} \frac{\sin y}{\pi y} = \frac{1}{\pi}\lim_{y \to 0}\frac{\sin y}{y} = \frac{1}{\pi} \]The limit is \(\displaystyle \dfrac{1}{\pi}\).

  6. Exercise 16

    lim⁡x→0cos⁡xπ−x\displaystyle \lim _{x \rightarrow 0} \frac{\cos x}{\pi-x}

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    NCERT’s answer
    $\displaystyle \frac{1}{\pi}$
    Direct substitution. This is not an indeterminate form: the denominator tends to \(\displaystyle \pi - 0 = \pi \neq 0\), and \(\displaystyle \cos x\) is continuous.\[\lim_{x \to 0} \frac{\cos x}{\pi - x} = \frac{\cos 0}{\pi - 0} = \frac{1}{\pi} \]The limit is \(\displaystyle \dfrac{1}{\pi}\).

  7. Exercise 17

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

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    NCERT’s answer
    $\displaystyle 4$
    Half-angle identity \(\displaystyle 1-\cos\theta = 2\sin^{2}\frac{\theta}{2}\).The form is \(\displaystyle \frac{0}{0}\). Rewrite both parts:\[\cos 2x - 1 = -2\sin^{2}x, \qquad \cos x - 1 = -2\sin^{2}\frac{x}{2} \]So, for \(\displaystyle x\) near \(\displaystyle 0\) but \(\displaystyle x \neq 0\),\[\frac{\cos 2x-1}{\cos x-1} = \frac{\sin^{2}x}{\sin^{2}\dfrac{x}{2}} = \frac{\left(2\sin\dfrac{x}{2}\cos\dfrac{x}{2}\right)^{2}}{\sin^{2}\dfrac{x}{2}} = 4\cos^{2}\frac{x}{2} \]using \(\displaystyle \sin x = 2\sin\frac{x}{2}\cos\frac{x}{2}\). Letting \(\displaystyle x \to 0\),\[\lim_{x \to 0} \frac{\cos 2x-1}{\cos x-1} = 4\cos^{2}0 = 4 \]The limit is \(\displaystyle 4\).

  8. Exercise 18

    lim⁡x→0ax+xcos⁡xbsin⁡x\displaystyle \lim _{x \rightarrow 0} \frac{a x+x \cos x}{b \sin x}

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    NCERT’s answer
    $\displaystyle \frac{a+1}{b}$
    Take out the common factor \(\displaystyle x\), then use \(\displaystyle \lim\limits_{x \to 0}\dfrac{x}{\sin x} = 1\).The form is \(\displaystyle \frac{0}{0}\). The numerator has \(\displaystyle x\) as a common factor:\[\frac{ax + x\cos x}{b\sin x} = \frac{x(a+\cos x)}{b\sin x} = \frac{1}{b}\,(a+\cos x)\cdot\frac{x}{\sin x} \]As \(\displaystyle x \to 0\), \(\displaystyle \dfrac{x}{\sin x} \to 1\) and \(\displaystyle \cos x \to 1\), so\[\lim_{x \to 0} \frac{ax + x\cos x}{b\sin x} = \frac{1}{b}(a+1)(1) = \frac{a+1}{b} \]The limit is \(\displaystyle \dfrac{a+1}{b}\).

  9. Exercise 19

    lim⁡x→0xsec⁡x\displaystyle \lim _{x \rightarrow 0} x \sec x

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    NCERT’s answer
    $\displaystyle 0$
    Direct substitution. Write \(\displaystyle \sec x = \dfrac{1}{\cos x} \), so\[x\sec x = \frac{x}{\cos x}. \]Both \(\displaystyle x\) and \(\displaystyle \cos x\) are continuous at \(\displaystyle 0\), and \(\displaystyle \cos 0 = 1 \neq 0\), so the quotient rule for limits applies and the limit is just the value:\[\lim_{x\to 0} x\sec x = \frac{\lim_{x\to 0} x}{\lim_{x\to 0}\cos x} = \frac{0}{1} = 0. \]Answer: \(\displaystyle 0\).

  10. Exercise 20

    lim⁡x→0sin⁡ax+bxax+sin⁡bxa,b,a+b≠0\displaystyle \lim _{x \rightarrow 0} \frac{\sin a x+b x}{a x+\sin b x} a, b, a+b \neq 0,

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    NCERT’s answer
    $\displaystyle 1$
    Divide numerator and denominator by \(\displaystyle x\), then use \(\displaystyle \lim_{\theta\to 0}\dfrac{\sin\theta}{\theta}=1 \).Both numerator and denominator tend to \(\displaystyle 0\), so the expression is of the \(\displaystyle \frac{0}{0} \) form; dividing through by \(\displaystyle x\) (legitimate, since \(\displaystyle x \neq 0\) as \(\displaystyle x \to 0\)) gives\[\frac{\sin ax + bx}{ax + \sin bx} = \frac{\dfrac{\sin ax}{x} + b}{a + \dfrac{\sin bx}{x}}. \]Now rewrite each sine quotient so that the angle matches the denominator:\[\frac{\sin ax}{x} = a\cdot\frac{\sin ax}{ax} \xrightarrow[x\to 0]{} a\cdot 1 = a, \qquad \frac{\sin bx}{x} = b\cdot\frac{\sin bx}{bx} \xrightarrow[x\to 0]{} b. \](Here \(\displaystyle a \neq 0\) and \(\displaystyle b \neq 0\), so \(\displaystyle ax\) and \(\displaystyle bx\) really do tend to \(\displaystyle 0\) through non‑zero values.) The denominator tends to \(\displaystyle a + b \neq 0\), so the quotient rule for limits is valid:\[\lim_{x\to 0}\frac{\sin ax + bx}{ax + \sin bx} = \frac{a+b}{a+b} = 1. \]Answer: \(\displaystyle 1\).