We read your photo as:

दो हजार बारह - (हिंदी अक्षरों को पढ़कर संख्याओं को विकल्पों में से पहचानकर लिखिए ।)

Languages

ScanSolve AI Answer

Answers

6 parts
21

12a\dfrac{1}{2\sqrt{a}}

22

13\dfrac{1}{3}

23
1
24

32\dfrac{3}{2}

25

16\dfrac{1}{6}

26

12\dfrac{1}{2}

Step-by-step solution

22 steps

Part 21· 4 steps

  1. Write the given limit.

    lim⁡x→ax−ax−a\lim_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x - a}

  2. Multiply numerator and denominator by the conjugate x+a\sqrt{x} + \sqrt{a}.

    lim⁡x→a(x−a)(x+a)(x−a)(x+a)=lim⁡x→ax−a(x−a)(x+a)\lim_{x \to a} \frac{(\sqrt{x} - \sqrt{a})(\sqrt{x} + \sqrt{a})}{(x - a)(\sqrt{x} + \sqrt{a})} = \lim_{x \to a} \frac{x - a}{(x - a)(\sqrt{x} + \sqrt{a})}

  3. Cancel the common factor x−ax - a (valid as x≠ax \neq a in the limit).

    lim⁡x→a1x+a\lim_{x \to a} \frac{1}{\sqrt{x} + \sqrt{a}}

  4. Substitute x=ax = a.

    1a+a=12a\frac{1}{\sqrt{a} + \sqrt{a}} = \frac{1}{2\sqrt{a}}

    \dfrac{1}{2\sqrt{a}}

Part 22

13\dfrac{1}{3}

  1. Rewrite tan⁡x=sin⁡xcos⁡x\tan x = \frac{\sin x}{\cos x}.

    lim⁡x→0tan⁡x−xx3=lim⁡x→0sin⁡xcos⁡x−xx3=lim⁡x→0sin⁡x−xcos⁡xx3cos⁡x\lim_{x \to 0} \frac{\tan x - x}{x^3} = \lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - x}{x^3} = \lim_{x \to 0} \frac{\sin x - x \cos x}{x^3 \cos x}

  2. Since lim⁡x→0cos⁡x=1\lim_{x \to 0} \cos x = 1, the limit is lim⁡x→0sin⁡x−xcos⁡xx3\lim_{x \to 0} \frac{\sin x - x \cos x}{x^3}.

  3. Use Taylor expansions around x=0x = 0

    sin⁡x=x−x36+o(x3),cos⁡x=1−x22+o(x4)\sin x = x - \frac{x^3}{6} + o(x^3), \quad \cos x = 1 - \frac{x^2}{2} + o(x^4)

  4. Compute xcos⁡xx \cos x

    xcos⁡x=x(1−x22+o(x4))=x−x32+o(x5)x \cos x = x \left(1 - \frac{x^2}{2} + o(x^4)\right) = x - \frac{x^3}{2} + o(x^5)

  5. Numerator sin⁡x−xcos⁡x\sin x - x \cos x

    (x−x36+o(x3))−(x−x32+o(x5))=−x36+x32+o(x3)=(12−16)x3+o(x3)=13x3+o(x3)\left(x - \frac{x^3}{6} + o(x^3)\right) - \left(x - \frac{x^3}{2} + o(x^5)\right) = -\frac{x^3}{6} + \frac{x^3}{2} + o(x^3) = \left(\frac{1}{2} - \frac{1}{6}\right)x^3 + o(x^3) = \frac{1}{3}x^3 + o(x^3)

  6. Divide by x3x^3

    lim⁡x→013x3+o(x3)x3=13\lim_{x \to 0} \frac{\frac{1}{3}x^3 + o(x^3)}{x^3} = \frac{1}{3}

    \dfrac{1}{3}

Part 231

  1. Substitute t=π2−xt = \frac{\pi}{2} - x.

    As x→π2x \to \frac{\pi}{2}, t→0t \to 0.

    lim⁡x→π2(π2−x)tan⁡x=lim⁡t→0ttan⁡(π2−t)=lim⁡t→0tcot⁡t\lim_{x \to \frac{\pi}{2}} \left(\frac{\pi}{2} - x\right) \tan x = \lim_{t \to 0} t \tan\left(\frac{\pi}{2} - t\right) = \lim_{t \to 0} t \cot t

  2. Rewrite cot⁡t=1tan⁡t\cot t = \frac{1}{\tan t}.

    tcot⁡t=ttan⁡t=tcos⁡tsin⁡tt \cot t = \frac{t}{\tan t} = \frac{t \cos t}{\sin t}

  3. lim⁡t→0tcos⁡tsin⁡t=lim⁡t→0(tsin⁡t⋅cos⁡t)=(lim⁡t→0tsin⁡t)(lim⁡t→0cos⁡t)=1⋅1=1\lim_{t \to 0} \frac{t \cos t}{\sin t} = \lim_{t \to 0} \left( \frac{t}{\sin t} \cdot \cos t \right) = \left( \lim_{t \to 0} \frac{t}{\sin t} \right) \left( \lim_{t \to 0} \cos t \right) = 1 \cdot 1 = 1

    1

Part 24

32\dfrac{3}{2}

  1. Factor the numerator using 1−u3=(1−u)(1+u+u2)1 - u^3 = (1 - u)(1 + u + u^2) where u=cos⁡xu = \cos x.

    1−cos⁡3x=(1−cos⁡x)(1+cos⁡x+cos⁡2x)1 - \cos^3 x = (1 - \cos x)(1 + \cos x + \cos^2 x) lim⁡x→01−cos⁡3xx2=lim⁡x→0(1−cos⁡x)(1+cos⁡x+cos⁡2x)x2=lim⁡x→0[1−cos⁡xx2⋅(1+cos⁡x+cos⁡2x)]\lim_{x \to 0} \frac{1 - \cos^3 x}{x^2} = \lim_{x \to 0} \frac{(1 - \cos x)(1 + \cos x + \cos^2 x)}{x^2} = \lim_{x \to 0} \left[ \frac{1 - \cos x}{x^2} \cdot (1 + \cos x + \cos^2 x) \right]

  2. Evaluate each limit separately.

    lim⁡x→01−cos⁡xx2=12,lim⁡x→0(1+cos⁡x+cos⁡2x)=1+1+12=3\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, \quad \lim_{x \to 0} (1 + \cos x + \cos^2 x) = 1 + 1 + 1^2 = 3

  3. Multiply the limits.

    12×3=32\frac{1}{2} \times 3 = \frac{3}{2}

    \dfrac{3}{2}

Part 25

16\dfrac{1}{6}

  1. Use Taylor expansion for sin⁡x\sin x around x=0x = 0.

    sin⁡x=x−x36+o(x3)\sin x = x - \frac{x^3}{6} + o(x^3)

  2. Numerator x−sin⁡xx - \sin x

    x−sin⁡x=x−(x−x36+o(x3))=x36+o(x3)x - \sin x = x - \left(x - \frac{x^3}{6} + o(x^3)\right) = \frac{x^3}{6} + o(x^3)

  3. Divide by x3x^3

    lim⁡x→0x−sin⁡xx3=lim⁡x→0x36+o(x3)x3=16\lim_{x \to 0} \frac{x - \sin x}{x^3} = \lim_{x \to 0} \frac{\frac{x^3}{6} + o(x^3)}{x^3} = \frac{1}{6}

    \dfrac{1}{6}

Part 26

12\dfrac{1}{2}

  1. Rewrite tan⁡x−sin⁡x\tan x - \sin x.

    tan⁡x−sin⁡x=sin⁡xcos⁡x−sin⁡x=sin⁡x(1cos⁡x−1)=sin⁡x⋅1−cos⁡xcos⁡x\tan x - \sin x = \frac{\sin x}{\cos x} - \sin x = \sin x \left( \frac{1}{\cos x} - 1 \right) = \sin x \cdot \frac{1 - \cos x}{\cos x} lim⁡x→0tan⁡x−sin⁡xx3=lim⁡x→0sin⁡x(1−cos⁡x)x3cos⁡x=lim⁡x→0[sin⁡xx⋅1−cos⁡xx2⋅1cos⁡x]\lim_{x \to 0} \frac{\tan x - \sin x}{x^3} = \lim_{x \to 0} \frac{\sin x (1 - \cos x)}{x^3 \cos x} = \lim_{x \to 0} \left[ \frac{\sin x}{x} \cdot \frac{1 - \cos x}{x^2} \cdot \frac{1}{\cos x} \right]

  2. Evaluate each limit separately.

    lim⁡x→0sin⁡xx=1,lim⁡x→01−cos⁡xx2=12,lim⁡x→01cos⁡x=1\lim_{x \to 0} \frac{\sin x}{x} = 1, \quad \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}, \quad \lim_{x \to 0} \frac{1}{\cos x} = 1

  3. Multiply the limits.

    1⋅12⋅1=121 \cdot \frac{1}{2} \cdot 1 = \frac{1}{2}

    \dfrac{1}{2}

Was this helpful?

Try one yourself

A new problem that tests the same skill.

Still stuck on this one?

Ask a follow-up, and the answer starts from this question.

Got a different question?Ask your own question

More Languages Questions

Qदो हजार बार - (हिंदी अक्षरों को पढ़कर संख्याओं को विकल्पों में से पहचानकर लिखिए 1) लोकगीतों ______ कई…Q10 अन्योक्ति अलंकार की परिभाषा उदाहरण सहित लिखिए। अथवा महाकाव्य किसे कहते हैं? दो महाकाव्यों के नाम…Qनीचे लिखे वाक्यों को पूरा कीजिए- 2. प्रत्येक स्तम्भ से शब्द लेकर वाक्य बनाइए और लिखिए-Qदिए गए मुहावरों के अर्थ और वाक्य प्रयोग लिखें।Qदिए गए शब्दों के विलोम शब्द लिखिए, मुहावरों के अर्थ लिखकर वाक्यों में प्रयोग कीजिए, और पाठ में आए…
See all Languages questions →
Got a syllabus? Photograph it and I'll tell you what's due and what's worth the most.