Limits And Derivatives CBSE Questions & Answers

\(\mathop {Lt}\limits_{x \to 0} \;{{\log (1 + a\;x) - \log (1 + b\;x)} \over x}\) is equal to

\(\mathop {Lt}\limits_{x \to 0} \;{{1 - \cos x} \over {\sqrt {1 + x} - 1}}\) is equal to

Let f (x) = x sin\({1 \over x}\), x\( \ne \)0, then the value of the function at x = 0, so that f is continuous at x = 0, is

\(\mathop {Lt}\limits_{x \to 0} \;\;{{\tan x - \sin x} \over {{x^3}}}\)is equal to

\(\mathop {Lt}\limits_{x \to 0} \;\;{{x\log (1 + x)} \over {1 - \cos x}}\) is equal to

\(\mathop {Lt}\limits_{x \to 0} \;\;{{{e^{px}} - {e^{qx}}} \over x}\) is equal to

\(\mathop {Lt}\limits_{x \to 0} \;\;{{{{(1 - x)}^n} - 1} \over x}\) is

\(\mathop {Lt}\limits_{x \to 0} \;\;{{\tan x - x} \over {{x^2}\tan x}}\)is equal to

\(\mathop {Lt}\limits_{x \to 0} \;\;\left[ {\cos x} \right]\)

\(\mathop {Lt}\limits_{x \to 0} \;\;{{\sin x - x} \over {{x^3}}}\) is equal to

\(\mathop {Lt}\limits_{x \to 0} \;\;\left( {{{\sin x - x} \over x}} \right)\cos \left( {{1 \over x}} \right)\) is equal to

\(\mathop {Lt}\limits_{x \to 0} \;\;\left( {{{\sin x - x} \over {{x^4}}}} \right)\) sin x equal to

Let \(f(x) = {{2 - {{(256 - 7x)}^{1/8}}} \over {{{(5x + 32)}^{1/5}} - 2}},x \ne 0,\) then for f to be continuous at x = 0, f (0) must be equal to

\(If(x) = \left\{ {\frac{{\sin \left[ x \right]}}{\begin{gathered} \left[ x \right] \\ 0,\;\;\;\; \\ \; \\ \end{gathered} },\mathop {\left[ x \right] \ne 0,then\;\mathop {Lt}\limits_{x \to 0} f(x)}\limits_{\left[ x \right] = 0} } \right.\)

\(\mathop {Lt}\limits_{x \to \infty } x\left( {{a^{1/x}} - 1} \right)\), where a > 0, is equal to