Limits, Continuity and Differentiability · Mathematics · WB JEE

$$\mathop {\lim }\limits_{x \to {\pi \over 2}} {{{a^{\cot x}} - {a^{\cos x}}} \over {\cot x - \cos x}},a > 0$$

Rolle's theorem is not applicable to the function $$f(x) = |x|$$ for $$ - 2 \le x \le 2$$ because

The value of the limit $$\mathop {\lim }\limits_{x \to 2} {{{e^{3x - 6}} - 1} \over {\sin (2 - x)}}$$ is

The $$\mathop {\lim }\limits_{x \to 2} {5 \over {\sqrt 2 - \sqrt x }}$$ is

A function f(x) is defined as follows for real x

$$f(x) = \left\{ {\matrix{ {1 - {x^2}} & , & {for\,x < 1} \cr 0 & , & {for\,x = 1} \cr {1 + {x^2}} & , & {for\,x > 1} \cr } } \right.$$

Then

Let $$f(x) = {{\sqrt {x + 3} } \over {x + 1}}$$, then the value of $$\mathop {\lim }\limits_{x \to - 3 - 0} f(x)$$ is

$$f(x) = x + |x|$$ is continuous for

The value of $$\mathop {\lim }\limits_{x \to 1} {{\sin ({e^{x - 1}} - 1)} \over {\log x}}$$ is

The value of $$\mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x + \cos x - 1} \over {{x^2}}}$$ is

The value of $$\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + 5{x^2}} \over {1 + 3{x^2}}}} \right)^{{1 \over {{x^2}}}}}$$ is

If $$f(5) = 7$$ and $$f'(5) = 7$$, then $$\mathop {\lim }\limits_{x \to 5} {{x\,f(5) - 5f(x)} \over {x - 5}}$$ is given by

If $$y = (1 + x)(1 + {x^2})(1 + {x^4})\,.....\,(1 + {x^{2n}})$$, then the value of $${\left( {{{dy} \over {dx}}} \right)_{x = 0}}$$ is

The value of f(0) so that the function $$f(x) = {{1 - \cos (1 - \cos x)} \over {{x^4}}}$$ is continuous everywhere is

In which of the following functions, Rolle's theorem is applicable?

The value of $$\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + ..... + {x^n} - n} \over {x - 1}}$$ is

$$\mathop {\lim }\limits_{x \to 0} {{\sin (\pi {{\sin }^2}x)} \over {{x^2}}} = $$

If the function $$f(x) = \left\{ {\matrix{ {{{{x^2} - (A + 2)x + A} \over {x - 2}},} & {for\,x \ne 2} \cr {2,} & {for\,x = 2} \cr } } \right.$$ is continuous at x = 2, then

$$f(x) = \left\{ {\matrix{ {[x] + [ - x],} & {when\,x \ne 2} \cr {\lambda ,} & {when\,x = 0} \cr } } \right.$$

If f(x) is continuous at x = 2, the value of $$\lambda$$ will be

For the function $$f(x) = {e^{\cos x}}$$, Rolle's theorem is

$$f(x) = \left\{ {\matrix{ {0,} & {x = 0} \cr {x - 3,} & {x > 0} \cr } } \right.$$

The function f(x) is

The function f(x) = ax + b is strictly increasing for all real x if

$$ \text { Let } f(x)=\left|\begin{array}{ccc} \cos x & x & 1 \\ 2 \sin x & x^3 & 2 x \\ \tan x & x & 1 \end{array}\right| \text {, then } \lim _\limits{x \rightarrow 0} \frac{f(x)}{x^2}= $$

If $$\alpha, \beta$$ are the roots of the equation $$a x^2+b x+c=0$$ then $$\lim _\limits{x \rightarrow \beta} \frac{1-\cos \left(a x^2+b x+c\right)}{(x-\beta)^2}$$ is

$$\mathop {\lim }\limits_{x \to \infty } \left\{ {x - \root n \of {(x - {a_1})(x - {a_2})\,...\,(x - {a_n})} } \right\}$$ where $${a_1},{a_2},\,...,\,{a_n}$$ are positive rational numbers. The limit

Let $$f:[1,3] \to R$$ be continuous and be derivable in (1, 3) and $$f'(x) = {[f(x)]^2} + 4\forall x \in (1,3)$$. Then

f(x) is a differentiable function and given $$f'(2) = 6$$ and $$f'(1) = 4$$, then $$L = \mathop {\lim }\limits_{h \to 0} {{f(2 + 2h + {h^2}) - f(2)} \over {f(1 + h - {h^2}) - f(1)}}$$

Let $$f(x) = \left\{ {\matrix{ {x + 1,} & { - 1 \le x \le 0} \cr { - x,} & {0 < x \le 1} \cr } } \right.$$

Let $$f(x) = [{x^2}]\sin \pi x,x > 0$$. Then

The value of $$\mathop {\lim }\limits_{n \to \infty } \left[ {\left( {{1 \over {2\,.\,3}} + {1 \over {{2^2}\,.\,3}}} \right) + \left( {{1 \over {{2^2}\,.\,{3^2}}} + {1 \over {{2^3}\,.\,{3^2}}}} \right)\, + \,...\, + \,\left( {{2 \over {{2^n}\,.\,{3^n}}} + {1 \over {{2^{n + 1}}\,.\,3n}}} \right)} \right]$$ is

The values of a, b, c for which the function $$f(x) = \left\{ \matrix{ {{\sin (a + 1)x + \sin x} \over x},x < 0 \hfill \cr c,x = 0 \hfill \cr {{{{(x + b{x^2})}^{{1 \over 2}}} - {x^{{1 \over 2}}}} \over {b{x^{{1 \over 2}}}}},x > 0 \hfill \cr} \right.$$ is continuous at x = 0, are

Let $$f(x) = {a_0} + {a_1}|x| + {a_2}|x{|^2} + {a_3}|x{|^3}$$, where $${a_0},{a_1},{a_2},{a_3}$$ are real constants. Then f(x) is differentiable at x = 0

$$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x}\ln \sqrt {{{1 + x} \over {1 - x}}} } \right)$$ is

Let f : [a, b] $$\to$$ R be continuous in [a, b], differentiable in (a, b) and f(a) = 0 = f(b). Then

$$\mathop {\lim }\limits_{x \to \infty } \left( {{{{x^2} + 1} \over {x + 1}} - ax - b} \right),(a,b \in R)$$ = 0. Then

If N = n! (n $$\in$$ N, n > 2), then find $$\mathop {\lim }\limits_{N \to \infty } \left[ {{{({{\log }_2}N)}^{ - 1}} + {{({{\log }_3}N)}^{ - 1}} + \,\,.....\,\, + {{({{\log }_n}N)}^{ - 1}}} \right]$$.

Use the formula $$\mathop {\lim }\limits_{x \to 0} {{{a^x} - 1} \over x} = {\log _e}a$$, to compute $$\mathop {\lim }\limits_{x \to 0} {{{2^x} - 1} \over {\sqrt {1 + x} - 1}}$$.

If f(a) = 2, f'(a) = 1, g(a) = $$-$$1 and g'(a) = 2, find the value of $$\mathop {\lim }\limits_{x \to a} {{g(x)f(a) - g(a)f(x)} \over {x - a}}$$.

Let R be the set of real numbers and f : R $$\to$$ R be such that for all x, y $$\in$$ R, $$|f(x) - f(y)| \le |x - y{|^3}$$. Prove that f is a constant function.