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NCERT Exemplar · Class 11 Mathematics Limits and Derivatives

80 questions · 80 still being checked

EXERCISE 13.3 1–10 (part 1 of 8)

  1. Evaluate :

    Exercise 1

    lim⁡x→3x2−9x−3\displaystyle \lim _{x \rightarrow 3} \frac{x^2-9}{x-3}

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    NCERT’s answer
    $\displaystyle 6$
    Factorise; the common factor cancels since \(\displaystyle x\neq 3\) in the limit.\[\frac{x^2-9}{x-3}=\frac{(x-3)(x+3)}{x-3}=x+3 \]\[\lim_{x\to3}(x+3)=6 \]Answer: \(\displaystyle 6\)

  2. Exercise 2

    lim⁡x→124x2−12x−1\displaystyle \lim _{x \rightarrow \frac{1}{2}} \frac{4 x^2-1}{2 x-1}

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    NCERT’s answer
    $\displaystyle 2$
    Factorise; the common factor cancels since \(\displaystyle x\neq \tfrac12\) in the limit.\[\frac{4x^2-1}{2x-1}=\frac{(2x-1)(2x+1)}{2x-1}=2x+1 \]\[\lim_{x\to\frac12}(2x+1)=2 \]Answer: \(\displaystyle 2\)

  3. Exercise 3

    lim⁡h→0x+h−xh\displaystyle \lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}

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

    Rationalise the numerator.\[\frac{\sqrt{x+h}-\sqrt{x}}{h}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=\frac{(x+h)-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)} \]\[=\frac{1}{\sqrt{x+h}+\sqrt{x}} \]\[\lim_{h\to0}\frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{2\sqrt{x}} \]Answer: \(\displaystyle \dfrac{1}{2\sqrt{x}}\) for \(\displaystyle x>0\)

  4. Exercise 4

    lim⁡x→0(x+2)13−213x\displaystyle \lim _{x \rightarrow 0} \frac{(x+2)^{\frac{1}{3}}-2^{\frac{1}{3}}}{x}

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    NCERT’s answer
    \(\displaystyle \frac{1}{3}2^{\frac{-2}{3}}\)
    Put \(\displaystyle y=x+2\); then \(\displaystyle y\to2\) as \(\displaystyle x\to0\), and \(\displaystyle x=y-2\).\[\lim_{x\to0}\frac{(x+2)^{\frac13}-2^{\frac13}}{x}=\lim_{y\to2}\frac{y^{\frac13}-2^{\frac13}}{y-2} \]\[\lim_{y\to a}\frac{y^{n}-a^{n}}{y-a}=n\,a^{n-1} \]\[=\frac13\cdot2^{\frac13-1}=\frac13\cdot2^{-\frac23} \]Answer: \(\displaystyle \dfrac{1}{3\cdot2^{2/3}}\)

  5. Exercise 5

    lim⁡x→1(1+x)6−1(1+x)2−1\displaystyle \lim _{x \rightarrow 1} \frac{(1+x)^6-1}{(1+x)^2-1}

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

    At \(\displaystyle x=1\) the denominator is not zero, so substitute directly.\[(1+x)^2-1\Big|_{x=1}=2^2-1=3\neq0 \]\[\lim_{x\to1}\frac{(1+x)^6-1}{(1+x)^2-1}=\frac{2^6-1}{2^2-1}=\frac{63}{3}=21 \]Answer: \(\displaystyle 21\)

  6. Exercise 6

    lim⁡x→a(2+x)52−(a+2)52x−a\displaystyle \lim _{x \rightarrow a} \frac{(2+x)^{\frac{5}{2}}-(a+2)^{\frac{5}{2}}}{x-a}

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    NCERT’s answer
    \(\displaystyle \frac{5}{2}(a+2)^{\frac{3}{2}}\)
    Put \(\displaystyle y=2+x\) and \(\displaystyle b=a+2\); then \(\displaystyle y\to b\) as \(\displaystyle x\to a\), and \(\displaystyle x-a=y-b\).\[\lim_{x\to a}\frac{(2+x)^{\frac52}-(a+2)^{\frac52}}{x-a}=\lim_{y\to b}\frac{y^{\frac52}-b^{\frac52}}{y-b} \]\[\lim_{y\to b}\frac{y^{n}-b^{n}}{y-b}=n\,b^{n-1} \]\[=\frac52\,b^{\frac32}=\frac52(a+2)^{\frac32} \]Answer: \(\displaystyle \dfrac52\,(a+2)^{\frac32}\)

  7. Exercise 7

    lim⁡x→1x4−xx−1\displaystyle \lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 7$
    Put \(\displaystyle y=\sqrt{x}\); then \(\displaystyle y\to1\) as \(\displaystyle x\to1\).\[\frac{x^4-\sqrt{x}}{\sqrt{x}-1}=\frac{y^8-y}{y-1}=\frac{y\,(y^7-1)}{y-1} \]\[\lim_{y\to1}\frac{y^7-1}{y-1}=7\cdot1^{6}=7 \]\[\lim_{y\to1}y\cdot\frac{y^7-1}{y-1}=1\cdot7=7 \]Answer: \(\displaystyle 7\)

  8. Exercise 8

    lim⁡x→2x2−43x−2−x+2\displaystyle \lim _{x \rightarrow 2} \frac{x^2-4}{\sqrt{3 x-2}-\sqrt{x+2}}

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 8$
    Rationalise the denominator.\[(\sqrt{3x-2}-\sqrt{x+2})(\sqrt{3x-2}+\sqrt{x+2})=(3x-2)-(x+2)=2(x-2) \]\[\frac{x^2-4}{\sqrt{3x-2}-\sqrt{x+2}}=\frac{(x-2)(x+2)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{2(x-2)}=\frac{(x+2)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{2} \]\[\lim_{x\to2}\frac{(x+2)\left(\sqrt{3x-2}+\sqrt{x+2}\right)}{2}=\frac{4\,(2+2)}{2}=8 \]Answer: \(\displaystyle 8\)

  9. Exercise 9

    lim⁡x→2x4−4x2+32x−8\displaystyle \lim _{x \rightarrow \sqrt{2}} \frac{x^4-4}{x^2+3 \sqrt{2 x}-8}

    Check this one against your book

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

    The denominator does not vanish at \(\displaystyle x=\sqrt2\), so substitute.\[x^4-4\to(\sqrt2)^4-4=0 \]\[x^2+3\sqrt{2x}-8\to 2+3\cdot2^{3/4}-8=3\cdot2^{3/4}-6\approx-0.955\neq0 \]\[\lim_{x\to\sqrt2}\frac{x^4-4}{x^2+3\sqrt{2x}-8}=\frac{0}{-0.955}=0 \]For the denominator \(\displaystyle x^2+3\sqrt2\,x-8\) (the \(\displaystyle 0/0\) form), factorise instead.\[\frac{x^4-4}{x^2+3\sqrt2\,x-8}=\frac{(x-\sqrt2)(x+\sqrt2)(x^2+2)}{(x-\sqrt2)(x+4\sqrt2)}\to\frac{2\sqrt2\cdot4}{5\sqrt2}=\frac85 \]Answer: \(\displaystyle 0\) as printed; \(\displaystyle \dfrac85\) for the denominator \(\displaystyle x^2+3\sqrt2\,x-8\)

  10. Exercise 10

    lim⁡x→1x7−2x5+1x3−3x2+2\displaystyle \lim _{x \rightarrow 1} \frac{x^7-2 x^5+1}{x^3-3 x^2+2}

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    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 1$
    Both numerator and denominator vanish at \(\displaystyle x=1\); divide out \(\displaystyle x-1\).\[x^7-2x^5+1=(x-1)(x^6+x^5-x^4-x^3-x^2-x-1) \]\[x^3-3x^2+2=(x-1)(x^2-2x-2) \]\[\lim_{x\to1}\frac{x^7-2x^5+1}{x^3-3x^2+2}=\lim_{x\to1}\frac{x^6+x^5-x^4-x^3-x^2-x-1}{x^2-2x-2} \]\[=\frac{1+1-1-1-1-1-1}{1-2-2}=\frac{-3}{-3}=1 \]Answer: \(\displaystyle 1\)