SATHEE: Limit Continuity and Differentiability 2 Question 6

Limit Continuity and Differentiability 2 Question 6

6. For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then,

$lim_{x \to 0^{+}} x \frac{1}{x} + \frac{2}{x} + \dots + \frac{15}{x}$

(2018 Main)

(a) is equal to 0

(b) is equal to 15

(d) does not exist (in $R$ )

Show Answer

Answer:

Correct Answer: 6. (b)

Solution:

$lim_{x \to 0} \frac{\int_{0}^{x^{2}} \cos^{2} t d t}{x \sin x}$

$\frac{0}{0}$ form

Applying L’Hospital’s rule, we get

$= lim_{x \to 0} \frac{\cos^{2} (x^{2}) \cdot 2 x - 0}{x \cos x + \sin x} = lim_{x \to 0} \frac{2 \cdot \cos^{2} (x^{2})}{\cos x + \frac{\sin x}{x}} = \frac{2}{1 + 1} = 1$

Limit Continuity and Differentiability 2 Question 6

6. For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then,

Answer:

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Limit Continuity and Differentiability 2 Question 6

6. For each t∈R, let [t] be the greatest integer less than or equal to t. Then,

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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6. For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then,