Limits, Continuity and Differentiability · Mathematics · KCET

Match the following:

In the following, $[\mathrm{x}]$ denotes the greatest integer less than or equal to x .

The function $f(x)=\left\{\begin{array}{ll}e^x+a x & , x<0 \\ b(x-1)^2 & , x \geq 0\end{array}\right.$ is differentiable at $x=0$. Then

$$ \text { A function } f(x)=\left\{\begin{array}{cl} \frac{e^{\frac{1}{x}}-1}{e^{\frac{1}{x}}+1}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right. $$

$\lim \limits_{x \rightarrow \frac{\pi}{4}} \frac{\sqrt{2} \cos x-1}{\cot x-1}$ is equal to

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

The function $f(x)=|\cos x|$ is

$$\lim _\limits{n \rightarrow \infty}\left(\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\frac{n}{n^2+3^2}+\ldots+\frac{1}{5 n}\right)= $$

If $$\lim _\limits{x \rightarrow 0} \frac{\sin (2+x)-\sin (2-x)}{x}=A \cos B$$, then the values of $$A$$ and $$B$$ respectively are

The function $$f(x)=\cot x$$ is discontinuous on every point of the set

If $$f(x)=\left\{\begin{array}{cc}x^2-1, & 0< x<2 \\ 2 x+3, & 2 \leq x<3\end{array}\right.$$,

the quadratic equation whose roots are $$\lim _\limits{x \rightarrow 2^{-}} f(x)$$ and $$\lim _\limits{x \rightarrow 2^{+}} f(x)$$ is

$$\lim _\limits{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}=$$

Consider the following statements

Statement 1 : $$\lim _\limits{x \rightarrow 1} \frac{a x^2+b x+c}{x^2+b x+a}$$ is 1

(where $$a+b+c \neq 0$$).

Statement 2 : $$\lim _\limits{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}$$ is $$\frac{1}{4}$$.

If $$f(x)=\left|\begin{array}{ccc}\cos x & 1 & 0 \\ 0 & 2 \cos x & 3 \\ 0 & 1 & 2 \cos x\end{array}\right|$$, then $$\lim _\limits{x \rightarrow \pi} f(x)$$ is equal to

At $$x=1$$, the function

$$f(x)=\left\{\begin{array}{cc} x^3-1, & 1< x < \infty \\ x-1, & -\infty< x \leq 1 \end{array}\right. \text { is }$$

The right hand and left hand limit of the function are respectively.

$$f(x)=\left\{\begin{array}{cc} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.$$

$$\lim _\limits{x \rightarrow 0}\left(\frac{\tan x}{\sqrt{2 x+4}-2}\right) \text { is equal to }$$

If $$f(x)=\left\{\begin{array}{cc}\frac{1-\cos K x}{x \sin x}, & \text { if } x \neq 0 \\ \frac{1}{2}, & \text { if } x=0\end{array}\right.$$ is continuous at $$x=0$$, then the value of $$K$$ is

If $$f(x)=\left\{\begin{array}{cl}\frac{\sin 3 x}{e^{2 x}-1} ; & x \neq 0 \\ k-2 ; & x=0\end{array}\right.$$ is continuous at $$x=0$$, then $$k=$$

$$\sum_\limits{r=1}^n(2 r-1)=x$$ then, $$ \lim _\limits{n \rightarrow \infty}\left[\frac{1^3}{x^2}+\frac{2^3}{x^2}+\frac{3^3}{x^2}+\ldots+\frac{n^3}{x^2}\right]=$$

Rolle's theorem is not applicable in which one of the following cases?

If $f(x)=\left\{\begin{array}{clc}\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x} & \text { if }-1 \leq x<0 \\ \frac{2 x+1}{x-1} & \text { if } 0 \leq x \leq 1\end{array}\right.$

is continuous at $x=0$, then the value of $k$ is

If $f(x)=\left\{\begin{array}{cl}\frac{\log _e x}{x-1} & ; x \neq 1 \\ k & ; x=1\end{array}\right.$

is continuous at $x=1$, then the value of $k$ is

$$The\,\,value\,\,of\,\,\mathop {\lim }\limits_{\theta \to 0} {{1 - \cos 4\theta } \over {1 - \cos 6\theta }}\,\,is$$