Limits, Continuity and Differentiability

If $f(x) = \begin{cases} ax + 7 & \text{if } x < 1 \\ 3x - 1 & \text{if } x = 1 \\ \dfrac{b}{x + 3} & \text{if } x > 1 \end{cases}$ is continuous at $x = 1$, then

If $\lim\limits_{x \to 3}\left(\dfrac{x^2 - ax - 3b}{x - 3}\right) = 5$, then $a + b = $

If $f(x) = \begin{cases} x^2 - 1 & \text{if } x \geq 2 \\ x + 1 & \text{if } x < 2 \end{cases}$, then $\lim\limits_{x \to 1} f(x) + \lim\limits_{x \to 2} f(x) = $

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



Column - I
Column - II


(a)


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x|x|
(i)
continuo...

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

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

$\lim _{x \rightarrow 1} \frac{x^4-\sqrt{x}}{\sqrt{x}-1}$ is

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

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

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

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

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

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

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

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

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

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...

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

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

The right hand and left hand limit of the function are respectively.
$$f(x)=\left\{\begin{array}{cc} \frac{e^{1 / x}-1}{e^{1 / x}+1}, & \text { if } x \neq 0 \\ 0, & \text { if } x=0 \end{array}\right.$$

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

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

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

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 \(k=\)

The value of $\lim \limits_{x \rightarrow 0} \frac{[x]}{x}$ is :

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

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

If $f(x)=\left\{\begin{array}{cll}k x^2 & \text { if } & x \leq 2 \\ 3 & \text { if } & x>2\end{array}\right.$ is continuous at $x=2$, then the value of $k$ is