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NCERT Solutions · Class 11 Mathematics Limits and Derivatives

73 exercises · 73 still being checked

EXERCISE 12.1 1–10 (part 1 of 7)

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

    Exercise 1

    \(\displaystyle \lim _{x \rightarrow 3} x+3\)

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    NCERT’s answer
    $\displaystyle 6$
    Direct substitution. A polynomial is continuous everywhere, so its limit at a point is simply its value there.\[\lim_{x \to 3} (x+3) = 3 + 3 = 6 \]The limit is \(\displaystyle 6\).

  2. Exercise 2

    \(\displaystyle \lim _{x \rightarrow \pi}\left(x-\frac{22}{7}\right)\)

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    NCERT’s answer
    $\displaystyle \left(\pi-\frac{22}{7}\right)$
    Direct substitution. \(\displaystyle x - \frac{22}{7}\) is a polynomial in \(\displaystyle x\), so the limit is its value at \(\displaystyle x = \pi\).\[\lim_{x \to \pi}\left(x - \frac{22}{7}\right) = \pi - \frac{22}{7} \]Note that \(\displaystyle \frac{22}{7}\) is only an approximation to \(\displaystyle \pi\), not \(\displaystyle \pi\) itself, so this difference is not zero.The limit is \(\displaystyle \pi - \dfrac{22}{7}\).

  3. Exercise 3

    \(\displaystyle \lim _{r \rightarrow 1} \pi r^{2}\)

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    NCERT’s answer
    $\displaystyle \pi$
    Direct substitution. \(\displaystyle \pi r^{2}\) is a polynomial in \(\displaystyle r\) (with \(\displaystyle \pi\) a constant), so substitute \(\displaystyle r = 1\).\[\lim_{r \to 1} \pi r^{2} = \pi (1)^{2} = \pi \]The limit is \(\displaystyle \pi\).

  4. Exercise 4

    \(\displaystyle \lim _{x \rightarrow 4} \frac{4 x+3}{x-2}\)

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    NCERT’s answer
    $\displaystyle \frac{19}{2}$
    Quotient rule for limits. If \(\displaystyle \lim g(x) \neq 0\), then \(\displaystyle \lim \dfrac{f(x)}{g(x)} = \dfrac{\lim f(x)}{\lim g(x)}\).Here the denominator tends to \(\displaystyle 4 - 2 = 2 \neq 0\), so substitution is legitimate:\[\lim_{x \to 4} \frac{4x+3}{x-2} = \frac{4(4)+3}{4-2} = \frac{19}{2} \]The limit is \(\displaystyle \dfrac{19}{2}\).

  5. Exercise 5

    \(\displaystyle \lim _{x \rightarrow-1} \frac{x^{10}+x^{5}+1}{x-1}\)

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    NCERT’s answer
    $\displaystyle -\frac{1}{2}$
    Quotient rule for limits. The denominator tends to \(\displaystyle -1-1 = -2 \neq 0\), so we may substitute directly.\[\lim_{x \to -1} \frac{x^{10}+x^{5}+1}{x-1} = \frac{(-1)^{10}+(-1)^{5}+1}{-1-1} = \frac{1-1+1}{-2} = -\frac{1}{2} \]The limit is \(\displaystyle -\dfrac{1}{2}\).

  6. Exercise 6

    \(\displaystyle \lim _{x \rightarrow 0} \frac{(x+1)^{5}-1}{x}\)

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    NCERT’s answer
    $\displaystyle 5$
    Substitution to the standard limit \(\displaystyle \lim\limits_{y \to a}\dfrac{y^{n}-a^{n}}{y-a} = n a^{\,n-1}\).Direct substitution gives the indeterminate form \(\displaystyle \frac{0}{0}\). Put \(\displaystyle y = x+1\); as \(\displaystyle x \to 0\), \(\displaystyle y \to 1\), and \(\displaystyle x = y-1\):\[\lim_{x \to 0} \frac{(x+1)^{5}-1}{x} = \lim_{y \to 1} \frac{y^{5}-1^{5}}{y-1} = 5 \cdot 1^{4} = 5 \](The same answer follows from expanding \(\displaystyle (x+1)^5 = 1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5\) and cancelling \(\displaystyle x\).)The limit is \(\displaystyle 5\).

  7. Exercise 7

    \(\displaystyle \lim _{x \rightarrow 2} \frac{3 x^{2}-x-10}{x^{2}-4}\)

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    NCERT’s answer
    $\displaystyle \frac{11}{4}$
    Factorise and cancel. At \(\displaystyle x = 2\) both numerator and denominator vanish, giving \(\displaystyle \frac{0}{0}\); so \(\displaystyle (x-2)\) is a factor of each.\[3x^{2}-x-10 = (x-2)(3x+5), \qquad x^{2}-4 = (x-2)(x+2) \]For \(\displaystyle x \neq 2\) we may cancel \(\displaystyle (x-2)\), and a limit at \(\displaystyle x \to 2\) never uses the value at \(\displaystyle x = 2\):\[\lim_{x \to 2} \frac{3x^{2}-x-10}{x^{2}-4} = \lim_{x \to 2} \frac{3x+5}{x+2} = \frac{11}{4} \]The limit is \(\displaystyle \dfrac{11}{4}\).

  8. Exercise 8

    \(\displaystyle \lim _{x \rightarrow 3} \frac{x^{4}-81}{2 x^{2}-5 x-3}\)

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    NCERT’s answer
    $\displaystyle \frac{108}{7}$
    Factorise and cancel. At \(\displaystyle x = 3\) both parts vanish, so \(\displaystyle (x-3)\) divides each.\[x^{4}-81 = (x^{2}-9)(x^{2}+9) = (x-3)(x+3)(x^{2}+9) \] \[2x^{2}-5x-3 = (x-3)(2x+1) \]Cancelling \(\displaystyle (x-3)\), which is non-zero for \(\displaystyle x \neq 3\):\[\lim_{x \to 3} \frac{x^{4}-81}{2x^{2}-5x-3} = \lim_{x \to 3} \frac{(x+3)(x^{2}+9)}{2x+1} = \frac{6 \times 18}{7} = \frac{108}{7} \]The limit is \(\displaystyle \dfrac{108}{7}\).

  9. Exercise 9

    \(\displaystyle \lim _{x \rightarrow 0} \frac{a x+b}{c x+1}\)

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    NCERT’s answer
    $\displaystyle b$
    Quotient rule for limits. The denominator tends to \(\displaystyle c(0)+1 = 1 \neq 0\), so substitute \(\displaystyle x = 0\).\[\lim_{x \to 0} \frac{ax+b}{cx+1} = \frac{a(0)+b}{c(0)+1} = b \]The limit is \(\displaystyle b\).

  10. Exercise 10

    \(\displaystyle \lim _{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\)

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    NCERT’s answer
    $\displaystyle 2$
    Standard limit \(\displaystyle \lim\limits_{z \to 1}\dfrac{z^{n}-1}{z-1} = n\) (valid for every real \(\displaystyle n\), being the case \(\displaystyle a = 1\) of \(\displaystyle \lim\limits_{z\to a}\frac{z^n-a^n}{z-a} = na^{\,n-1}\)).The form is \(\displaystyle \frac{0}{0}\). Divide numerator and denominator by \(\displaystyle z-1\), which is non-zero for \(\displaystyle z \neq 1\):\[\lim_{z \to 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1} = \lim_{z \to 1} \frac{\dfrac{z^{\frac{1}{3}}-1}{z-1}}{\dfrac{z^{\frac{1}{6}}-1}{z-1}} = \frac{\dfrac{1}{3}}{\dfrac{1}{6}} = 2 \](Equivalently, put \(\displaystyle z = t^{6}\) with \(\displaystyle t \to 1\): the expression becomes \(\displaystyle \frac{t^{2}-1}{t-1} = t+1 \to 2\).)The limit is \(\displaystyle 2\).