Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Continuity and Differentiability

Question:

The value of $\underset{n→∞}{\lim}\frac{(x+1)^{10}+(x+2)^{10}+...+(x+100)^{10}}{x^{10}+10^{10}}$ is:

Options:

1

0

100

50

Correct Answer:

100

Explanation:

$\underset{n→∞}{\lim}\frac{(x+1)^{10}+(x+2)^{10}+...+(x+100)^{10}}{x^{10}+10^{10}}$

$=\underset{n→∞}{\lim}\frac{x^{10}[(1+\frac{1}{x})^{10}+(1+\frac{2}{x})^{10}+...+(1+\frac{100}{x})^{10}]}{x^{10}[1+\frac{10^{10}}{x^{10}}]}=1+1+1+...+1=100$