Question 515 Marks
Discuss the continuity of the following functions at the indicated point:
$\text{f}\text{(x)}=\begin{cases}\frac{{1}-\text{x}^\text{n}}{1-\text{x}}, & \text{x} \neq1\\\text{n}-1, & \text{ x} = 1\end{cases}\text{ n }\in\ \text{N at x}=1$
AnswerWe want, to check the continuity at x = 1
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 1^-}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0} \text{f}(1-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{1-(1-\text{h})^\text{n}}{1-(1-\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{1-\Big[1-\text{nh}+\frac{\text{n}(\text{n}-1)}{2}\text{h}^2+\dots\Big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{n}-\frac{\text{n(n-1)}}{2!}\text{h}+\dots$
$=\text{n}$
$\text{RHL}=\lim\limits_{\text{x} \rightarrow 1^+}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0} \text{f}\text{(1+h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{1-(1+\text{h})^\text{n}}{1-(1+\text{h})}=\lim\limits_{\text{h} \rightarrow 0}\frac{1-\Big[1-\text{nh}+\frac{\text{n}(\text{n}-1)}{2}\text{h}^2+\dots\Big]}{\text{h}}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{n}+\frac{\text{n}(\text{n}-1)}{2!}\text{h}+\dots$
$=\text{n}$
$\text{f}(1)=\text{n}-1$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}( 1)$
Hence, funtion is discontinuous at x = 1
This is removable discotinuity.
View full question & answer→Question 525 Marks
Find the value of k if f(x) is continuous at $\text{x}=\frac{\pi}{2},$ where
$\text{f}\text{(x)}=\begin{cases}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}, &\text{ x}\neq\frac{\pi}{2}\\3, &\text{ x}=\frac{\pi}{2}\end{cases}$
AnswerSince f(x) is continuous at $\text{x}=\frac{\pi}{2},$ L.H.Limit = R.H.Limit.
$\Rightarrow\lim\limits_{\text{x} \rightarrow \frac{\pi^-}{2}}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \frac{\pi^+}{2}}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\text{f}\text{(x)}=\text{f}\Big(\frac{\pi}{2}\Big)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow \frac{\pi^-}{2}}\frac{\text{k}\cos\text{x}}{\pi-2\text{x}}=3$
$\Rightarrow\text{k}\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{2\Big(\frac{\pi}{2}-\text{x}\Big)}=3$
$\Rightarrow\frac{\text{k}}{2}\lim\limits_{\text{x} \rightarrow \frac{\pi}{2}}\frac{\sin\Big(\frac{\pi}{2}-\text{x}\Big)}{2\Big(\frac{\pi}{2}-\text{x}\Big)}=3$
$\Rightarrow\frac{\text{k}}{2}=3$
$\Rightarrow\text{k}=6$
View full question & answer→Question 535 Marks
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}\text{at x} = \pi$
AnswerGiven,
$\text{f(x)}=\begin{cases}\text{k}+1,&\text{if}\text{ x}\leq\pi\\\cos\text{x},&\text{if}\text{ x}>\pi\end{cases}$
We have,
$(\text{LHL at x}= \pi)=\lim_\limits{\text{x}\rightarrow\pi^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(\pi-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\text{k}(\pi-\text{h})+1=\text{k}\pi+1$
$(\text{RHL at x}= \pi)=\lim_\limits{\text{x}\rightarrow\pi^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(\pi+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\cos(\pi+\text{h})=\cos\pi=-1$
If f(x) is continuous at $\text{x}=\pi,$ then
$\lim_\limits{\text{x}\rightarrow\pi^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\pi^+}\text{f(x)}$
$\Rightarrow\text{k}\pi+1=-1$
$\Rightarrow\text{k}=\frac{-2}{\pi}$
View full question & answer→Question 545 Marks
Discuss the continuity of the function $\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
AnswerWhen x < 2, we have
f(x) = 2x - 1
We know that a polynomial function is everywhere continuous.
So, f(x) is continuous for each x < 2
When x > 2, we have
$\text{f(x)}=\frac{3\text{x}}{2}$
The functions 3x and 2 are continuous being the polynomial and constant function, respectively.
Thus, the quotient function $\frac{3\text{x}}{2}$ is continuous at each x > 2
Now,
Let us consider the point x = 2
$\text{f(x)}=\begin{cases}2\text{x}-1,&\text{if }\text{ x}<2\\\frac{3\text{x}}{2},&\text{if }\text{ x}\geq2\end{cases}$
We have
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2-\text{h})\\=\lim\limits_{\text{h}\rightarrow0}(2(2-\text{h})-1)=4-1=3$
$(\text{LHL at x}= 2)=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}(2+\text{h})\\=\lim\limits_{\text{h}\rightarrow0}\frac{3(\text{h}+2)}{2}=3$
Also,
$\text{f}(2)=\frac{3(2)}{2}=3$
$\therefore\ \lim\limits_{\text{x}\rightarrow2^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow2^+}\text{f(x)}=\text{f}(2)$
Thus, f(x) is continuous at x = 2
Hence, f(x) is everywhere continuous.
View full question & answer→Question 555 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\sin\text{x}-\cos\text{x},&\text{if }\text{ x}\neq0\\-1,&\text{if }\text{ x}=0\end{cases}$
AnswerThe given function f is $\text{f(x)}=\begin{cases}\sin\text{x}-\cos\text{x},&\text{if }\text{ x}\neq0\\-1,&\text{if }\text{ x}=0\end{cases}$ It is evident that f is defind at all points of the real line. Let c be a real number.Case I:
If $\text{c}\neq0,$ then $\text{f(c)}=\sin\text{c}-\cos\text{c}$ $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(\sin\text{x}-\cos\text{x})=\sin\text{c}-\cos\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that $\text{x}\neq0$Case II:
If c = 0, then f(0) = -1 $\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0}(\sin\text{x}-\cos\text{x})\\=\sin0-\cos0=0-1=-1$ $\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0}(\sin\text{x}-\cos\text{x})\\=\sin0-\cos0=0-1=-1$ Therefore, f is continuousb at x = 0 From tha above observations, it can be continuous that f is continuous at every point of the real line. Thus, f is a continuous function.
View full question & answer→Question 565 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}\frac{\sin\text{x}}{\text{x}},&\text{if }\text{ x}<0\\2\text{x}+3,&\text{ x}\geq0\end{cases}$
AnswerWhen x < 0, we have $\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$
We know that $\sin\text{x}$ and the identity function continuous for x < 0, so the quotient function
$\text{f(x)}=\frac{\sin\text{x}}{\text{x}}$ is continuous for x < 0
When x > 0 f(x) = 2x + 3, which is a polynomial of degree 1. So, f(x) 2x + 3 is continuous for x > 0
Now, consider the point x = 0
$\text{LHL}=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(-\text{h})}{-\text{h}}=\lim_\limits{\text{h}\rightarrow0}\frac{-\sin\text{h}}{-\text{h}}=1$
$\text{RHL}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\frac{\sin(\text{h})}{\text{h}}=1$
$\text{f}(0)=2\times0+3=3$
Thus, $\text{LHL}=\text{RHL}\neq\text{f}(0)$
Hence, f(x) is discontinuous at x = 0
View full question & answer→Question 575 Marks
If $\text{f(x)}=\frac{\tan\big(\frac{\pi}{4}-\text{x}\big)}{\cot2\text{x}}$ for $\text{x}\neq\frac{\pi}{4},$ find the value which can be assigned to f(x) at $\text{x}=\frac{\pi}{4}$ so that the function f(x) becomes continuous every where in $\Big[0,\frac{\pi}{2}\Big]$
AnswerWhere $\text{x}\neq\frac{\pi}{4},\ \tan\big(\frac{\pi}{4}-\text{x}\big)$ and $\cot2\text{x}$ are continuous in $\Big[0,\frac{\pi}{2}\Big]$
Thus, the quotient function $\frac{\tan\Big(\frac{\pi}{4}-\text{x}\Big)}{\cot2\text{x}}$ is continuous in $\Big[0,\frac{\pi}{2}\Big]$ for each $\text{x}\neq\frac{\pi}{4}$
So, if f(x) is continuous at $\text{x}=\frac{\pi}{4},$ then it will be everywhere continuous in $\Big[0,\frac{\pi}{2}\Big]$
Now,
Let us consider the point $\text{x}=\frac{\pi}{4}$
Given, $\text{f(x)}=\frac{\tan\big(\frac{\pi}{4}-\text{x}\big)}{\cot2\text{x}},\text{x}\neq\frac{\pi}{4}$
We have
$\Big(\text{LHL at x}=\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\pi}{4}-\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\tan\big(\frac{\pi}{4}-\frac{\pi}{4}+\text{h}\big)}{\cot\big(\frac{\pi}{2}-2\text{h}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(\text{h})}{\tan(2\text{h})}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\frac{\tan(\text{h})}{\text{h}}}{\frac{2\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}\Bigg(\frac{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(\text{h})}{\text{h}}}{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}$
$\Big(\text{RHL at x}=\frac{\pi}{4}\Big)=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^+}\text{f(x)}=\lim\limits_{\text{h}\rightarrow0}\text{f}\Big(\frac{\pi}{4}+\text{h}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\tan\big(\frac{\pi}{4}-\frac{\pi}{4}-\text{h}\big)}{\cot\big(\frac{\pi}{2}+2\text{h}\big)}\Bigg)=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(-\text{h})}{-\tan(2\text{h})}\Big)\\=\lim\limits_{\text{h}\rightarrow0}\Big(\frac{\tan(\text{h})}{\tan(2\text{h})}\Big)=\lim\limits_{\text{h}\rightarrow0}\Bigg(\frac{\frac{\tan(\text{h})}{\text{h}}}{\frac{2\tan(2\text{h})}{2\text{h}}}\Bigg)\\=\frac{1}{2}\Bigg(\frac{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(\text{h})}{\text{h}}}{\lim\limits_{\text{h}\rightarrow0}\frac{\tan(2\text{h})}{2\text{h}}}\Bigg)=\frac{1}{2}$
If f(x) is continuous at $\text{x}=\frac{\pi}{4},$ then
$\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\lim\limits_{\text{x}\rightarrow\frac{\pi}{4}^-}\text{f(x)}=\text{f}\Big(\frac{\pi}{4}\Big)$
$\therefore\ \text{f}\Big(\frac{\pi}{4}\Big)=\frac{1}{2}$
Hence, for $\text{f}\Big(\frac{\pi}{4}\Big)=\frac{1}{2},$ the function f(x) will be everywhere continuous in $\Big[0,\frac{\pi}{2}\Big]$
View full question & answer→Question 585 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{2^\text{z+2}-16}{4^\text{x}-16}, &\text{if x} \neq 2\\\text{k}, & \text{x} = 2\end{cases}$
is continuous at x = 2, Find k.
AnswerWe are given that the function is continuous at x = 2
$\therefore$ LHL = RHL = f(2)...(i)
Now,
f(2) = k...(A)
$\text{LHL}=\lim\limits_{\text{x} \rightarrow 2^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(2-\text{h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{(2-\text{h)+2}}-16}{4^{(2-\text{h)}}-16}=\lim\limits_{\text{h} \rightarrow 0}\frac{2^{2-\text{h}}-16}{4^{2-\text{h}}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^4-2^\text{-h}-16}{4^2.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16.2^\text{-h}-16}{16.4^\text{-h}-16}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{16\Big(2^\text{-h}-1\Big)}{16\Big(4^\text{-h}-1\Big)}$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}\Big)-1^2}$ $\Big[\because2^{-2\text{h}}=\Big(2^{-\text{h}}\Big)^2=4^{-\text{h}}\Big]$
$=\lim\limits_{\text{h} \rightarrow 0}\frac{2^\text{-h}-1}{\Big(2^\text{-h}-1\Big)\Big(2^\text{-h}+1\Big)}=\frac{1}{2}\dots(\text{B})$
$\therefore$ Using (i) from (A) & (B)
$\text{k}=\frac{1}{2}$
View full question & answer→Question 595 Marks
Discuss the continuity of the f(x) at the indicated points f(x) = |x| + |x - 1| at x = 0, 1.
AnswerGiven,
$\text{f(x)}=|\text{x}|+|\text{x}-1|$
We have,
$(\text{LHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0-\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|0-\text{h}|+|0-\text{h}-1|\big]=1$
$(\text{RHL at x}= 0)=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{h}\rightarrow0}\text{f}(0+\text{h})$
$=\lim_\limits{\text{h}\rightarrow0}\big[|0+\text{h}|+|0+\text{h}-1|\big]=1$
Also, $\text{f}(1)=|1|+|1-1|=1+0=1$
$\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\text{f}(0)$ and $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^+}\text{f(x)}=\text{f}=1$
Hence, f(x) is continuous at x = 0, 1
View full question & answer→Question 605 Marks
Find the points of discontinuity, if any of the following function:
$\text{f(x)}=\begin{cases}2\text{x},&\text{ if}\text{ x}<0\\0,&\text{if }0\leq\text{x}\leq1\\4\text{x},&\text{if }\text{ x}>1\end{cases}$
AnswerThe given function is $\text{f(x)}=\begin{cases}2\text{x},&\text{ if}\text{ x}<0\\0,&\text{if }0\leq\text{x}\leq1\\4\text{x},&\text{if }\text{ x}>1\end{cases}$ The given function is defined at all points of the real line. Let c be a point on the real line.Case I:
If c < 0 then f(c) = 2c $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(2\text{x})=2\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuous at all points x, such that x < 0Case II:
If c = 0, then f(c) = f(0) = 0 The left hand limit of f at x = 0 is, $\lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^-}(2\text{x})=2\times0=0$ The right hand limit of at x = 0 is, $\lim_\limits{\text{x}\rightarrow0^+}\text{f(x)}=\lim_\limits{\text{x}\rightarrow0^+}(0)=0$ $\therefore\ \lim_\limits{\text{x}\rightarrow0^-}\text{f(x)}=\text{f}(0)$ Therefore, f is continuous at x = 0Case III:
If 0 < c < 1, then f(x) = 0 and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}(0)=0$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f}(\text{c})$ Therefore, if is continuous at all points of the interval (0, 1)Case IV:
If c = 1, then f(c) = f(1) = 0 The left hand limit of at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^-}(0)=0$ The right hand limit of f at x = 1 is, $\lim_\limits{\text{x}\rightarrow1^-}\text{f(x)}=\lim_\limits{\text{x}\rightarrow1^-}(4\text{x})=4\times1=4$ It is observes that the left and right hand limits of f at x = 1 do not coincide. Therefore, f is not continuous at x = 1Case V:
If c < 1, then f(c) = 4c and $\lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\lim_\limits{\text{x}\rightarrow\text{c}}\text{(4x)}=4\text{c}$ $\therefore\ \lim_\limits{\text{x}\rightarrow\text{c}}\text{f(x)}=\text{f(c)}$ Therefore, f is continuouse at all points x such that x > 1 Hence, f is not continuous only at x = 1
View full question & answer→Question 615 Marks
If $\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}, & \text{x} \neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$ is continuous at x = 0. find k.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}, & \text{x} \neq 0\\\frac{1}{2}, & \text{x}= 0\end{cases}$
If f(x) is continuous at x = 0, then
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\text{f}(0)\dots(\text{i})$
Consider:
$\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\Bigg(\frac{1-\cos\text{kx}}{\text{x}\sin\text{x}}\Bigg)=\lim\limits_{\text{x} \rightarrow 0}\Bigg(\frac{2\sin^2\frac{\text{kx}}{2}}{\text{x}\sin\text{x}}\Bigg)$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{2\sin^2\frac{\text{kx}}{2}}{\text{x}^2\Big(\frac{\sin\text{x}}{\text{x}}\Big)}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow 0}\begin{pmatrix}\frac{\frac{2\text{k}^2}{4}\Big(\sin\frac{\text{kx}}{\text{x}}\Big)^2}{\Big(\frac{\text{kx}}{2}\Big)^2\Big(\frac{\sin\text{x}}{\text{x}}\Big)}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2\text{k}^2}{4}\begin{pmatrix}\frac{\lim\limits_{\text{x} \rightarrow 0}\frac{\Big(\sin\frac{\text{kx}}{2}\Big)^2}{\Big(\frac{\text{kx}}{2}\Big)^2}}{\lim\limits_{\text{x} \rightarrow 0}\frac{\sin\text{x}}{\text{x}}}\end{pmatrix}$
$\Rightarrow\lim\limits_{\text{x} \rightarrow 0}\text{f}\text{(x)}=\frac{2\text{k}^2}{4}\times1=\frac{\text{k}^2}{2}$
From equation (i), we have
$\frac{\text{k}^2}{2}=\text{f}(0)$
$\Rightarrow\frac{\text{k}^2}{2}=\frac{1}{2}$
$\Rightarrow\text{k}=\pm1$
View full question & answer→Question 625 Marks
If $\text{f}\text{(x)}=\begin{cases}\text{e}^\frac{1}{\text {x}}, & \text{if} \text{ x}\neq 0\\1, & \text{if}\text{x} = 0\end{cases}$ find whethe f is continuous at x = 0.
AnswerGiven,
$\text{f}\text{(x)}=\begin{cases}\text{e}^\frac{1}{\text {x}}, & \text{if} \text{ x}\neq 0\\1, & \text{if}\text{x} = 0\end{cases}$
We observe
$(\text{LHL at x}=0)=\lim\limits_{\text{x} \rightarrow 0^-}\text{f}\text{(x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}(0-\text{x)}=\lim\limits_{\text{h} \rightarrow 0}\text{f}\text{(-h)}$
$=\lim\limits_{\text{h} \rightarrow 0}\text{e}^\frac{-1}{\text{h}}=\lim\limits_{\text{h} \rightarrow 0}\bigg(\frac{1}{\text{e}^{\frac{1}{\text{h}}}}\bigg)=\frac{1}{\lim\limits_{\text{h} \rightarrow 0}\text{e}^{\frac{1}{\text{h}}}}=0$
$(\text{RHL at x}=0)=\lim\limits_{\text{h} \rightarrow 0^+}\text{f}(\text{x})=\lim\limits_{\text{h} \rightarrow 0}\text{f}(\text{h})$
$=\lim\limits_{\text{h} \rightarrow 0^+}\text{e}^\frac{1}{\text{h}}=\infty$
Given,
$\text{f}(0)=1$
It is known for a function f(x) to be continuous at x = a,
$\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}=\lim\limits_{\text{x} \rightarrow \text{a}^+}\text{f}\text{(x)}=\text{f}\text{(a)}$
But here,
$\lim\limits_{\text{x} \rightarrow \text{a}^-}\text{f}\text{(x)}\neq\lim\limits_{\text{x} \rightarrow \text{0}^+}\text{f}\text{(x)}$
Hence, f(x) is discontinuous at x = 0.
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