SolveItClass 11 · NCERT

NCERT Solutions · Class 11 Mathematics Limits and Derivatives

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EXERCISE 12.1 21–32 (part 3 of 7)

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

    Exercise 21

    lim⁡x→0(cosec⁡x−cot⁡x)\displaystyle \lim _{x \rightarrow 0}(\operatorname{cosec} x-\cot x)

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    NCERT’s answer
    $\displaystyle 0$
    Combine into one fraction, then use the half‑angle identities.\[\operatorname{cosec} x - \cot x = \frac{1}{\sin x} - \frac{\cos x}{\sin x} = \frac{1-\cos x}{\sin x}. \]This is of the \(\displaystyle \frac{0}{0} \) form, so simplify first. Using \(\displaystyle 1-\cos x = 2\sin^2\dfrac{x}{2} \) and \(\displaystyle \sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2} \),\[\frac{1-\cos x}{\sin x} = \frac{2\sin^{2}\dfrac{x}{2}}{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}} = \tan\frac{x}{2} \qquad (\sin\tfrac{x}{2}\neq 0 \text{ near } x=0,\ x\neq 0). \]Since \(\displaystyle \tan \) is continuous at \(\displaystyle 0\),\[\lim_{x\to 0}\left(\operatorname{cosec} x - \cot x\right) = \lim_{x\to 0}\tan\frac{x}{2} = \tan 0 = 0. \]Answer: \(\displaystyle 0\).

  2. Exercise 22

    lim⁡x→π2tan⁡2xx−π2\displaystyle \lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 x}{x-\dfrac{\pi}{2}}

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    Shift the variable so that the limit point becomes \(\displaystyle 0\).Put \(\displaystyle y = x - \dfrac{\pi}{2} \), so \(\displaystyle x = y + \dfrac{\pi}{2} \) and \(\displaystyle x \to \dfrac{\pi}{2} \) becomes \(\displaystyle y \to 0 \). Then\[\tan 2x = \tan\!\left(2y + \pi\right) = \tan 2y, \]because \(\displaystyle \tan \) has period \(\displaystyle \pi \). Hence\[\lim_{x\to \frac{\pi}{2}} \frac{\tan 2x}{x - \dfrac{\pi}{2}} = \lim_{y\to 0}\frac{\tan 2y}{y} = \lim_{y\to 0} 2\cdot\frac{\tan 2y}{2y} = 2\cdot 1 = 2, \]using the standard limit \(\displaystyle \lim_{\theta\to 0}\dfrac{\tan\theta}{\theta} = 1 \).Answer: \(\displaystyle 2\).

  3. Exercise 23

    Find lim⁡x→0f(x)\displaystyle \lim _{x \rightarrow 0} f(x) and lim⁡x→1f(x)\displaystyle \lim _{x \rightarrow 1} f(x), where f(x)={2x+3,x≤03(x+1),x>0\displaystyle f(x)=\left\{\begin{array}{cc}2 x+3, & x \leq 0 \\ 3(x+1), & x>0\end{array}\right.

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    One‑sided limits at the break point; direct substitution elsewhere.The definition changes at \(\displaystyle x = 0\), so at that point the left‑hand and right‑hand limits must be computed separately and compared.At \(\displaystyle x = 0\). For \(\displaystyle x < 0\) the rule is \(\displaystyle 2x+3\), and for \(\displaystyle x > 0\) it is \(\displaystyle 3(x+1)\):\[\lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} (2x+3) = 3, \qquad \lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 3(x+1) = 3(0+1) = 3. \]The two agree, so the limit exists and \(\displaystyle \lim_{x\to 0} f(x) = 3 \).At \(\displaystyle x = 1\). Every point sufficiently close to \(\displaystyle 1\) satisfies \(\displaystyle x > 0\), so only the rule \(\displaystyle 3(x+1)\) is used from both sides:\[\lim_{x\to 1} f(x) = 3(1+1) = 6. \]Answer: \(\displaystyle \lim_{x\to 0} f(x) = 3 \) and \(\displaystyle \lim_{x\to 1} f(x) = 6 \).

  4. Exercise 24

    Find lim⁡x→1f(x)\displaystyle \lim _{x \rightarrow 1} f(x), where f(x)={x2−1,x≤1−x2−1,x>1\displaystyle f(x)= \begin{cases}x^{2}-1, & x \leq 1 \\ -x^{2}-1, & x>1\end{cases}

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    NCERT’s answer
    Limit does not exist at $\displaystyle x=1$
    Compare the one‑sided limits at the break point.The rule changes at \(\displaystyle x = 1\), so test both sides.\[\lim_{x\to 1^{-}} f(x) = \lim_{x\to 1^{-}} \left(x^{2}-1\right) = 1 - 1 = 0, \]\[\lim_{x\to 1^{+}} f(x) = \lim_{x\to 1^{+}} \left(-x^{2}-1\right) = -1 - 1 = -2. \]Since \(\displaystyle 0 \neq -2 \), the left‑hand and right‑hand limits are different.Answer: \(\displaystyle \lim_{x\to 1} f(x) \) does not exist.

  5. Exercise 25

    Evaluate lim⁡x→0f(x)\displaystyle \lim _{x \rightarrow 0} f(x), where f(x)={∣x∣x,x≠00,x=0\displaystyle f(x)= \begin{cases}\frac{|x|}{x}, & x \neq 0 \\ 0, & x=0\end{cases}

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    NCERT’s answer
    Limit does not exist at $\displaystyle x=0$
    Remove the modulus by splitting into \(\displaystyle x>0\) and \(\displaystyle x<0\).Recall \(\displaystyle |x| = x \) for \(\displaystyle x > 0 \) and \(\displaystyle |x| = -x \) for \(\displaystyle x < 0 \). So for \(\displaystyle x \neq 0\),\[\frac{|x|}{x} = \begin{cases} \dfrac{x}{x} = 1, & x > 0 \\[4pt] \dfrac{-x}{x} = -1, & x < 0 \end{cases} \]Therefore\[\lim_{x\to 0^{+}} f(x) = \lim_{x\to 0^{+}} 1 = 1, \qquad \lim_{x\to 0^{-}} f(x) = \lim_{x\to 0^{-}} (-1) = -1. \]The one‑sided limits are unequal. (The value \(\displaystyle f(0) = 0\) is irrelevant — a limit never looks at the value at the point itself.)Answer: \(\displaystyle \lim_{x\to 0} f(x) \) does not exist.

  6. Exercise 26

    Find lim⁡x→0f(x)\displaystyle \lim _{x \rightarrow 0} f(x), where f(x)={x∣x∣,x≠00,x=0\displaystyle f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 0, & x=0\end{array}\right.

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    NCERT’s answer
    Limit does not exist at $\displaystyle x=0$
    Remove the modulus by splitting into \(\displaystyle x>0\) and \(\displaystyle x<0\).For \(\displaystyle x \neq 0\), using \(\displaystyle |x| = x \) when \(\displaystyle x > 0 \) and \(\displaystyle |x| = -x \) when \(\displaystyle x < 0 \),\[\frac{x}{|x|} = \begin{cases} \dfrac{x}{x} = 1, & x > 0 \\[4pt] \dfrac{x}{-x} = -1, & x < 0 \end{cases} \]Hence\[\lim_{x\to 0^{+}} f(x) = 1, \qquad \lim_{x\to 0^{-}} f(x) = -1. \]The two one‑sided limits differ, and the value \(\displaystyle f(0)=0\) has no bearing on the limit.Answer: \(\displaystyle \lim_{x\to 0} f(x) \) does not exist.

  7. Exercise 27

    Find lim⁡x→5f(x)\displaystyle \lim _{x \rightarrow 5} f(x), where f(x)=∣x∣−5\displaystyle f(x)=|x|-5

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    NCERT’s answer
    $\displaystyle 0$
    Resolve the modulus on a neighbourhood of the limit point, then substitute.We are approaching \(\displaystyle x = 5\), and every \(\displaystyle x\) close to \(\displaystyle 5\) is positive, so \(\displaystyle |x| = x \) throughout that neighbourhood. Thus near \(\displaystyle x=5\),\[f(x) = |x| - 5 = x - 5, \]which is a polynomial and therefore continuous. Both one‑sided limits are the same:\[\lim_{x\to 5^{-}} f(x) = \lim_{x\to 5^{+}} f(x) = 5 - 5 = 0. \]Answer: \(\displaystyle \lim_{x\to 5} f(x) = 0 \).

  8. Exercise 28

    Suppose f(x)={a+bx,x<14,x=1b−ax,x>1\displaystyle f(x)= \begin{cases}a+b x, & x<1 \\ 4, & x=1 \\ b-a x, & x>1\end{cases} and if lim⁡x→1f(x)=f(1)\displaystyle \lim _{x \rightarrow 1} f(x)=f(1) what are possible values of a\displaystyle a and b\displaystyle b ?

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    NCERT’s answer
    $\displaystyle a=0, b=4$
    Force the two one‑sided limits to agree with \(\displaystyle f(1)\).\(\displaystyle \lim_{x\to 1} f(x) \) exists and equals \(\displaystyle f(1)\) exactly when the left‑hand limit, the right‑hand limit and \(\displaystyle f(1)\) are all equal. Here \(\displaystyle f(1) = 4\).Left‑hand limit (rule \(\displaystyle a+bx\) for \(\displaystyle x<1\)):\[\lim_{x\to 1^{-}} f(x) = a + b(1) = a + b. \]Right‑hand limit (rule \(\displaystyle b-ax\) for \(\displaystyle x>1\)):\[\lim_{x\to 1^{+}} f(x) = b - a(1) = b - a. \]So the conditions are\[a + b = 4 \qquad \text{and} \qquad b - a = 4. \]Adding the two equations: \(\displaystyle 2b = 8\), so \(\displaystyle b = 4\). Substituting back, \(\displaystyle a + 4 = 4\), so \(\displaystyle a = 0\).Check. With \(\displaystyle a=0,\ b=4\): left‑hand limit \(\displaystyle =0+4=4\), right‑hand limit \(\displaystyle =4-0=4\), and \(\displaystyle f(1)=4\). All three agree.Answer: \(\displaystyle a = 0\) and \(\displaystyle b = 4\).

  9. Exercise 29

    Let a1,a2,…,an\displaystyle a_{1}, a_{2}, \ldots, a_{n} be fixed real numbers and define a function f(x)=(x−a1)(x−a2)…(x−an).f(x)=\left(x-a_{1}\right)\left(x-a_{2}\right) \ldots\left(x-a_{n}\right) . What is lim⁡x→a1f(x)\displaystyle \lim _{x \rightarrow a_{1}} f(x) ? For some a≠a1,a2,…,an\displaystyle a \neq a_{1}, a_{2}, \ldots, a_{n}, compute lim⁡x→af(x)\displaystyle \lim _{x \rightarrow a} f(x).

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    NCERT’s answer
    $\displaystyle \lim _{x \rightarrow a_{1}} f(x)=0$ and $\displaystyle \lim _{x \rightarrow a} f(x)=\left(a-a_{1}\right)\left(a-a_{2}\right) \ldots\left(a-a_{x}\right)$
    A polynomial is continuous, so its limit is its value.\(\displaystyle f(x) = (x-a_{1})(x-a_{2})\cdots(x-a_{n}) \) is a product of \(\displaystyle n\) polynomials, hence a polynomial. By the product rule for limits (applied \(\displaystyle n-1\) times) together with \(\displaystyle \lim_{x\to c}(x - a_{i}) = c - a_{i} \),\[\lim_{x\to c} f(x) = (c-a_{1})(c-a_{2})\cdots(c-a_{n}) = f(c) \quad \text{for every real } c. \]Taking \(\displaystyle c = a_{1}\). The very first factor becomes \(\displaystyle a_{1}-a_{1}=0\), so the whole product is \(\displaystyle 0\):\[\lim_{x\to a_{1}} f(x) = (a_{1}-a_{1})(a_{1}-a_{2})\cdots(a_{1}-a_{n}) = 0. \]Taking \(\displaystyle c = a\) with \(\displaystyle a \neq a_{1}, a_{2}, \ldots, a_{n}\).\[\lim_{x\to a} f(x) = (a-a_{1})(a-a_{2})\cdots(a-a_{n}). \]Here no factor vanishes, so this limit is a non‑zero number.Answer: \(\displaystyle \lim_{x\to a_{1}} f(x) = 0 \), and \(\displaystyle \lim_{x\to a} f(x) = (a-a_{1})(a-a_{2})\cdots(a-a_{n}) \).

  10. Exercise 30

    If f(x)={∣x∣+1,x<00,x=0∣x∣−1,x>0\displaystyle f(x)=\left\{\begin{array}{ll}|x|+1, & x<0 \\ 0, & x=0 \\ |x|-1, & x>0\end{array}\right.. For what value (s) of a\displaystyle a does lim⁡x→af(x)\displaystyle \lim _{x \rightarrow a} f(x) exists?

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    NCERT’s answer
    $\displaystyle \lim _{x \rightarrow a} f(x)$ exists for all $\displaystyle a \neq 0$.
    Test the single break point separately from all other points.At \(\displaystyle a = 0\). For \(\displaystyle x<0\) we have \(\displaystyle |x| = -x\), so \(\displaystyle f(x) = -x+1\); for \(\displaystyle x>0\) we have \(\displaystyle |x| = x\), so \(\displaystyle f(x) = x-1\). Hence\[\lim_{x\to 0^{-}} f(x) = -0+1 = 1, \qquad \lim_{x\to 0^{+}} f(x) = 0-1 = -1. \]These differ, so \(\displaystyle \lim_{x\to 0} f(x) \) does not exist.At \(\displaystyle a < 0\). Every point near \(\displaystyle a\) is negative, so near \(\displaystyle a\) the function is the single polynomial \(\displaystyle f(x) = -x+1\), and\[\lim_{x\to a} f(x) = -a+1 = |a| + 1, \]which exists.At \(\displaystyle a > 0\). Every point near \(\displaystyle a\) is positive, so near \(\displaystyle a\) the function is \(\displaystyle f(x) = x-1\), and\[\lim_{x\to a} f(x) = a-1 = |a| - 1, \]which exists.Answer: \(\displaystyle \lim_{x\to a} f(x) \) exists for every \(\displaystyle a \neq 0\), and fails to exist only at \(\displaystyle a = 0\).

  11. Exercise 31

    If the function f(x)\displaystyle f(x) satisfies lim⁡x→1f(x)−2x2−1=π\displaystyle \lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1}=\pi, evaluate lim⁡x→1f(x)\displaystyle \lim _{x \rightarrow 1} f(x).

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    NCERT’s answer
    $\displaystyle 2$
    Recover \(\displaystyle f(x)-2\) as a product of two factors whose limits are known.For \(\displaystyle x \neq 1\) we may write the identity\[f(x) - 2 = \frac{f(x)-2}{x^{2}-1}\cdot\left(x^{2}-1\right). \]Both factors on the right have limits as \(\displaystyle x \to 1\): the first tends to \(\displaystyle \pi\) (given), and the second tends to \(\displaystyle 1^{2}-1 = 0\). Since both limits exist, the product rule for limits applies:\[\lim_{x\to 1}\left(f(x)-2\right) = \pi \times 0 = 0. \]Finally, \(\displaystyle f(x) = 2 + \left(f(x)-2\right) \), so by the sum rule\[\lim_{x\to 1} f(x) = 2 + 0 = 2. \]Answer: \(\displaystyle \lim_{x\to 1} f(x) = 2 \).

  12. Exercise 32

    If f(x)={mx2+n,x<0nx+m,0≤x≤1nx3+m,x>1\displaystyle f(x)=\left\{\begin{array}{ll}m x^{2}+n, & x<0 \\ n x+m, & 0 \leq x \leq 1 \\ n x^{3}+m, & x>1\end{array}\right.. For what integers m\displaystyle m and n\displaystyle n does both lim⁡x→0f(x)\displaystyle \lim _{x \rightarrow 0} f(x) and lim⁡x→1f(x)\displaystyle \lim _{x \rightarrow 1} f(x) exist?

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    NCERT’s answer
    For $\displaystyle \lim _{x \rightarrow 0} f(x)$ to exists, we need $\displaystyle m=n ; \lim _{x \rightarrow 1} f(x)$ exists for any integral value of $\displaystyle m$ and $\displaystyle n$.
    Match the one‑sided limits at each of the two break points.The definition changes at \(\displaystyle x = 0\) and at \(\displaystyle x = 1\); test each.At \(\displaystyle x = 0\). For \(\displaystyle x<0\) the rule is \(\displaystyle mx^{2}+n\); for \(\displaystyle x\) just above \(\displaystyle 0\) the middle rule \(\displaystyle nx+m\) applies.\[\lim_{x\to 0^{-}} f(x) = m(0)^{2}+n = n, \qquad \lim_{x\to 0^{+}} f(x) = n(0)+m = m. \]So \(\displaystyle \lim_{x\to 0} f(x) \) exists if and only if \(\displaystyle m = n \).At \(\displaystyle x = 1\). For \(\displaystyle x\) just below \(\displaystyle 1\) the middle rule \(\displaystyle nx+m\) applies; for \(\displaystyle x>1\) the rule is \(\displaystyle nx^{3}+m\).\[\lim_{x\to 1^{-}} f(x) = n(1)+m = m+n, \qquad \lim_{x\to 1^{+}} f(x) = n(1)^{3}+m = m+n. \]These are equal whatever \(\displaystyle m\) and \(\displaystyle n\) are, so \(\displaystyle \lim_{x\to 1} f(x) = m+n \) always exists — this break point imposes no condition.Combining, the only requirement is \(\displaystyle m = n\).Answer: both limits exist precisely when \(\displaystyle m = n\); that is, for any integer \(\displaystyle n\), take \(\displaystyle m = n\) (and then \(\displaystyle \lim_{x\to 0} f(x) = n \) and \(\displaystyle \lim_{x\to 1} f(x) = 2n \)).