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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Limits

NTA Abhyas 2020 · Mathematics · Limits

The value of _ x → ∞ 2 (x) ^ 1 / 2 + 3 (x) ^ 1 / 3 + 4 (x) ^ 1 / 4...

The value of lim⁡x→∞2(x)1/2+3(x)1/3+4(x)1/4+.....+(x)1/n(2x−3)1/2+(2x−3)1/3+....+(2x−3)1/n\lim_{x \to \infty} \dfrac{2{\left(x\right)}^{1 / 2} + 3{\left(x\right)}^{1 / 3} + 4{\left(x\right)}^{1 / 4} +..... + {\left(x\right)}^{1 / n}}{{\left(2x - 3\right)}^{1 / 2} + {\left(2x - 3\right)}^{1 / 3} +.... + {\left(2x - 3\right)}^{1 / n}} is

Options

  1. A2\sqrt{2}
  2. B22
  3. C13\dfrac{1}{\sqrt{3}}
  4. D00

Correct answer

A. 2\sqrt{2}

Step-by-step solution

Given,

lim⁡x→∞2x1/2+3x1/3+....+nx1/n(2x−3)1/2+(2x−3)1/3+...+(2x−3)1/n\lim_{x \to \infty} \dfrac{2{x}^{1 / 2} + 3{x}^{1 / 3} + . . . . + n{x}^{1 / n}}{{\left(2x - 3\right)}^{1 / 2} + {\left(2x - 3\right)}^{1 / 3} + . . . + {\left(2x - 3\right)}^{1 / n}}

=lim⁡h→02(1h1/2)+3(1h1/3)+....+n(1h1/n)1h1/2(2−3h)1/2+1h1/3(2−3h)1/3+.....+1h1/n(2−3h)1/n= \lim_{h \to 0} \dfrac{2\left(\dfrac{1}{{h}^{1 / 2}}\right) + 3\left(\dfrac{1}{{h}^{1 / 3}}\right) + . . . . + n\left(\dfrac{1}{{h}^{1 / n}}\right)}{\dfrac{1}{{h}^{1 / 2}}{\left(2 - 3h\right)}^{1 / 2} + \dfrac{1}{{h}^{1 / 3}}{\left(2 - 3h\right)}^{1 / 3} + . . . . . + \dfrac{1}{{h}^{1 / n}}{\left(2 - 3h\right)}^{1 / n}}

\left[Onputtingx = \dfrac{1}{h}asx \to \infty, h \to 0\right]<br/><br/><br /><br />= \lim_{h \to 0} \dfrac{2 + 3{h}^{\left(\dfrac{1}{2} - \dfrac{1}{3}\right)} + 4{h}^{\left(\dfrac{1}{2} - \dfrac{1}{4}\right)} + . . . . + n{h}^{\left(\dfrac{1}{2} - \dfrac{1}{n}\right)}}{{\left(2 - 3h\right)}^{1 / 2} + {h}^{\left(\dfrac{1}{2} - \dfrac{1}{3}\right)}{\left(2 - 3h\right)}^{1 / 3} + . . . . + {h}^{\left(\dfrac{1}{2} - \dfrac{1}{n}\right)}{\left(2 - 3h\right)}^{1 / n}}<br/><br/><br /><br />= \dfrac{2 + 0 + 0 + 0 + . . . . . . .}{{2}^{1 / 2} + 0 + 0 + . . . . . . .} = \sqrt{2} .

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