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

If the sum of an infinite decreasing G.P. is 3 and the sum of the squares of its term is \[\frac{9}{2}\], then write its first term and common difference.

\[\lim_{x \to 1} \frac{x^2 + 1}{x + 1}\] 

\[\lim_{x \to 0} \frac{2 x^2 + 3x + 4}{x^2 + 3x + 2}\] 

\[\lim_{x \to 3} \frac{\sqrt{2x + 3}}{x + 3}\] 

\[\lim_{x \to 1} \frac{\sqrt{x + 8}}{\sqrt{x}}\] 

\[\lim_{x \to a} \frac{\sqrt{x} + \sqrt{a}}{x + a}\] 

\[\lim_{x \to 1} \frac{1 + \left( x - 1 \right)^2}{1 + x^2}\]

\[\lim_{x \to 0} \frac{x^{2/3} - 9}{x - 27}\]

\[\lim_{x \to 0} 9\] 

\[\lim_{x \to 2} \left( 3 - x \right)\] 

\[\lim_{x \to - 1}{\left( 4 x^2 + 2 \right)}\]

\[\lim_{x \to - 1} \frac{x^3 - 3x + 1}{x - 1}\]

\[\lim_{x \to 0} \frac{3x + 1}{x + 3}\] 

\[\lim_{x \to 3} \frac{x^2 - 9}{x + 2}\] 

\[\lim_{x \to 0} \frac{ax + b}{cx + d}, d \neq 0\]

\[\lim_{x \to - 5} \frac{2 x^2 + 9x - 5}{x + 5}\] 

\[\lim_{x \to 3} \frac{x^2 - 4x + 3}{x^2 - 2x - 3}\] 

\[\lim_{x \to 3} \frac{x^4 - 81}{x^2 - 9}\] 

\[\lim_{x \to 2} \frac{x^3 - 8}{x^2 - 4}\] 

\[\lim_{x \to - 1/2} \frac{8 x^3 + 1}{2x + 1}\]