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 is
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Given,
\left[Onputtingx = \dfrac{1}{h}asx \to \infty, h \to 0\right]= \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}}= \dfrac{2 + 0 + 0 + 0 + . . . . . . .}{{2}^{1 / 2} + 0 + 0 + . . . . . . .} = \sqrt{2} .
\left[Onputtingx = \dfrac{1}{h}asx \to \infty, h \to 0\right]= \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}}= \dfrac{2 + 0 + 0 + 0 + . . . . . . .}{{2}^{1 / 2} + 0 + 0 + . . . . . . .} = \sqrt{2} .