Commerce (English Medium) कक्षा ११ - CBSE Question Bank Solutions

\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . . n^3}{n^4} \right]\]

\[\lim_{n \to \infty} \left[ \frac{1^3 + 2^3 + . . . n^3}{\left( n - 1 \right)^4} \right]\] 

\[\lim_{x \to \infty} \left[ \sqrt{x}\left\{ \sqrt{x + 1} - \sqrt{x} \right\} \right]\] 

\[\lim_{n \to \infty} \left[ \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + . . . + \frac{1}{3^n} \right]\] 

\[\lim_{x \to \infty} \left[ \frac{x^4 + 7 x^3 + 46x + a}{x^4 + 6} \right]\] where a is a non-zero real number. 

\[f\left( x \right) = \frac{a x^2 + b}{x^2 + 1}, \lim_{x \to 0} f\left( x \right) = 1\] and \[\lim_{x \to \infty} f\left( x \right) = 1,\]then prove that f(−2) = f(2) = 1

Show that \[\lim_{x \to \infty} \left( \sqrt{x^2 + x + 1} - x \right) \neq \lim_{x \to \infty} \left( \sqrt{x^2 + 1} - x \right)\] 

\[\lim_{x \to - \infty} \left( \sqrt{4 x^2 - 7x} + 2x \right)\] 

\[\lim_{x \to - \infty} \left( \sqrt{x^2 - 8x} + x \right)\] 

Evaluate: \[\lim_{n \to \infty} \frac{1^4 + 2^4 + 3^4 + . . . + n^4}{n^5} - \lim_{n \to \infty} \frac{1^3 + 2^3 + . . . + n^3}{n^5}\] 

Evaluate: \[\lim_{n \to \infty} \frac{1 . 2 + 2 . 3 + 3 . 4 + . . . + n\left( n + 1 \right)}{n^3}\] 

\[\lim_{x \to 0} \frac{\sin 3x}{5x}\] 

\[\lim_{x \to 0} \frac{\sin x^0}{x}\] 

\[\lim_{x \to 0} \frac{x^2}{\sin x^2}\] 

\[\lim_{x \to 0} \left[ \frac{x^2}{\sin x^2} \right]\] 

\[\lim_{x \to 0} \frac{\sin x \cos x}{3x}\] 

\[\lim_{x \to 0} \frac{3 \sin x - 4 \sin^3 x}{x}\] 

\[\lim_{x \to 0} \frac{\tan 8x}{\sin 2x}\] 

\[\lim_{x \to 0} \frac{\tan mx}{\tan nx}\] 

\[\lim_{x \to 0} \frac{\sin 5x}{\tan 3x}\]