Question

Answer the below Question !!

$1)\tt { \: \: if \: f(x) \: = \begin{cases}m {x}^{2} + n,x < 0 \\ nx + m,0 \leqslant x \leqslant 1 \\ {nx}^{3} + m,x > 1 \end{cases}}$
$\bf \: For \: \: what \: \: values \: \: of \: \: integers \: \: m \: and \: n \: does \: the \: limits \: \lim_{x \to0} \: f(x) \: and \: \lim_{x \to 1} \: f(x) \: exist .$
$2)\tt \: if \: f(x) = \begin{cases} |x| + 1 ,\: x < 0 \\ 0 \: \: \: \: \: \: ,\: x = 0 \\ |x| + 1 \: ,x > 0 \end{cases} \: \:$
$\bf \: For \: \: what values \: \: (s) \: \: of \: \underline{a} \: \: does \: \: \lim_{x\to a} \: \: f(x) \: \: exist ?$
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Answer 1

ANSWER 1:

• (i) f(x) is valid when m = n for the limits x → 0

• (ii) f(x) is valid for any value of m and n for the limits x → 1

GIVEN:

$\tt {If \ f(x) = \begin{cases} \tt m {x}^{2} + n,x < 0 \\  \tt nx + m,0 \leqslant x \leqslant 1 \\ \tt {nx}^{3} + m,x > 1 \end{cases}}$

TO FIND:

• The values of m and n for the given cases.

EXPLANATION:

$\tt {If \ f(x) = \begin{cases} \tt m {x}^{2} + n,x < 0 \\  \tt nx + m,0 \leqslant x \leqslant 1 \\ \tt {nx}^{3} + m,x > 1 \end{cases}} \\  \\  \\ \dashrightarrow \displaystyle\tt\lim_{x \to0} f(x) \ has \ L.H.L \ and \ R.H.L \\  \\  \\ \tt\dashrightarrow L.H.L \implies\displaystyle\tt\lim_{x \to{0}^{ - } } f(x) = m {x}^{2}  + n \\  \\  \\ \tt\dashrightarrow f(0) = m {(0)}^{2}  + n \\  \\  \\  \tt\dashrightarrow f(0) = n \\  \\  \\ \dashrightarrow \tt R.H.L \implies\displaystyle\tt\lim_{x \to{0}^{  +  } } f(x) = n {x}  + m \\  \\  \\ \tt\dashrightarrow f(0) = n {(0)}  + m\\  \\  \\  \tt\dashrightarrow f(0) = m \\  \\  \\  \tt\dashrightarrow Limit\ exists\ if \ L.H.L = R.H.L \\  \\  \\ \dashrightarrow \displaystyle\tt\lim_{x \to0} f(x) \ exists \ if \ m = n$

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$\displaystyle \tt\dashrightarrow \lim_{x \to1} f(x) \ has \ L.H.L\ and \ R.H.L \\  \\  \\ \displaystyle\tt\dashrightarrow L.H.L \implies\lim_{x \to{1}^{ - } } f(x) = nx + m \\  \\  \\ \dashrightarrow\tt f(0) = n {(1)}  + m \\  \\  \\  \dashrightarrow\tt f(1) = m + n \\  \\  \\ \dashrightarrow\tt R.H.L \implies\displaystyle\tt\lim_{x \to{1}^{ + } } f(x) = n{x}^{3}   + m \\  \\  \\ \dashrightarrow \tt f(1) = n {(1)}^{3}  + m\\  \\  \\  \tt f(0) = m + n \\  \\  \\ \dashrightarrow \tt Limit\ exists\ if \ L.H.L = R.H.L \\  \\  \\ \displaystyle\dashrightarrow \tt \lim_{x \to1} f(x) \ exists \ for \ any \ values \ of \ m \ and\ n.$

ANSWER 2:

• f(x) is valid when a ≠ 0 for the limits x → a,

GIVEN:

$\tt If \ f(x) = \begin{cases} \tt |x| + 1 ,\: x < 0 \\ \tt0 \: \: \: \: \: \: ,\: x = 0 \\ \tt |x| + 1 \: ,x > 0 \end{cases}$

TO FIND:

• The values of a for the given case.

EXPLANATION:

$\tt If \ f(x) = \begin{cases} \tt |x| + 1 ,\: x < 0 \\ \tt0 \: \: \: \: \: \: ,\: x = 0 \\ \tt |x| + 1 \: ,x > 0 \end{cases} \\  \\  \\ \dashrightarrow \displaystyle\tt\lim_{x \to a} f(x) \ has \ L.H.L \ and \ R.H.L \\  \\  \\ \dashrightarrow\tt when\ a=0\\ \\ \\ \dashrightarrow\tt L.H.L \implies\displaystyle\tt\lim_{x \to{0}^{ - } } f(x) = |x| + 1 \\  \\  \\ \dashrightarrow\tt f(0) = 1\\  \\  \\ \dashrightarrow\tt R.H.L \implies\displaystyle\tt\lim_{x \to{0}^{  +  } } f(x) =|x|-1 \\  \\  \\ \dashrightarrow\tt f(0) = -1\\  \\  \\  \dashrightarrow\tt Limit\ exists\ if \ L.H.L = R.H.L \\  \\  \\ \dashrightarrow \tt L.H.L \not= R.H.L\\ \\ \\ \dashrightarrow \displaystyle\tt \lim_{x \to a} f(x) \ doesnot\ exists \ if \ a = 0. \\ \\ \\ \dashrightarrow \tt When \ a < 0 \ f(a) = |a| + 1 \ is \ continuous. \\ \tt \:so \ limit \ exists. \\ \\ \\ \dashrightarrow\tt When \ a > 0\ f(a) = |a| - 1 \ is \ continuous. \\ \tt so \ limit \ exists. \\ \\ \\ \dashrightarrow\displaystyle\tt \lim_{x \to a} f(x) \ exists \ when \ a \ne 0$

Answer 2

Step-by-step explanation:

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${ \underline{ \underline{ \bf{Question:}}}}$

$1)\tt { \: \: if \: f(x) \: = \begin{cases}m {x}^{2} + n,x < 0 \\ nx + m,0 \leqslant x \leqslant 1 \\ {nx}^{3} + m,x > 1 \end{cases}}$

$\bf \: For \: \: what \: \: values \: \: of \: \: integers \: \: m \: and \: n \: does \: the \: limits \: \lim_{x \to0} \: f(x) \: and \: \lim_{x \to 1} \: f(x) \: exist .$

${ \underline{ \underline{ \bf{Answer :}}}}$

It is given that ,

${ \underline{ \boxed{ \pink{ \bf{\lim_{x\to 0} \: \: f(x) \: and \: \: \lim_{x \to 1} \: \: f(x) \: \: both \: \: exist \: }}}}}$

$\iff \bf \lim_{x \to \: {0}^{ - } } \: f(x) = \lim_{x \to \: {0}^{ + }} \: f(x) \: and \: \: \lim_{x \to \: {1}^{ - } } \: f(x) = \lim_{x \to \: {1}^{ + }} f(x)$

$\iff \bf \: \lim_{h \to \: 0}f(0 - h) = \lim_{h \to \: 0}f(0 + h) \: \: and \: \: \lim_{h \to \: 0}f(1 + h) = \lim_{h \to \: 0}f(1 + h)$

$\iff \bf \: \lim_{h \to \: 0} \: m {( - h)}^{2} + n = \lim_{h \to \: 0} \: n(h) + m \: \: and \: \lim_{h \to \: 0} \: n(1 - h) + m \: = \lim_{h \to \: 0}n( {1 + {h})^{3} } + m$

$\iff \bf \: n = m \: \: and \: n + m \: = n + m$

$\iff \bf \: m = n$

${ \underline{\rm{ \blue{Hence , \: \lim_{x\to 0} \: f(x) \: and \: \lim_{x\to 1} \: f(x) \: both \: sides \: for \: n=m}}} }$

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${ \underline{ \underline{ \bf{Question:}}}}$

$2)\tt \: if \: f(x) = \begin{cases} |x| + 1 ,\: x < 0 \\ 0 \: \: \: \: \: \: ,\: x = 0 \\ |x| + 1 \: ,x > 0 \end{cases} \: \:$

$\bf \: For \: \: what values \: \: (s) \: \: of \: \underline{a} \: \: does \: \: \lim_{x\to a} \: \: f(x) \: \: exist ?$

${ \underline{ \underline{ \bf{Answer:}}}}$

We have,

$\bf \: \: \: f(x) = \begin{cases} |x| + 1 ,\: x < 0 \\ 0 \: \: \: \: \: \: ,\: x = 0 \\ |x| + 1 \: ,x > 0 \end{cases} \: \:$

$\implies \: \bf \: \: \: f(x) = \begin{cases} - x + 1 ,\: x < 0 \\ 0 \: \: \: \: \: \: ,\: x = 0 \\ x+ 1 \: ,x > 0 \end{cases} \: \:$

$\qquad \qquad \qquad \qquad \qquad\qquad \qquad \qquad \qquad \qquad \: \bigg[\because |x| = \begin{cases}x,x>0 \\ -x,x<0 \end{cases} \bigg]$

$\rm \: Clearly , \: \: \lim_{x \to a} \: f(x) \: exist \: for \: all \: a≠ 0. \: So \: let \: us \: see \: whether \: \lim_{x \to 0} \: f(x) \: exist \: or \: not .$

we observe that,

$\iff\bf \lim_{x \to 0^-} f(x) = \lim_{x \to 0} f(0-h) = \lim_{h \to 0} -(-h) +1=1$

and,

$\iff\bf \lim_{x\to 0^+} f(x) =\lim_{h \to 0} f(0+h)= \lim_{h \to 0} h-1=-1$

$\underline{\boxed{\bf{\green{\therefore \lim_{x \to 0^-}f(x) ≠ \lim_{x \to 0^+} f(x)}}}}$

So,

$\underline{ \rm{ \blue{\lim_{x \to 0} \: f(x) \: does \: not \: exist \: . \: Hence \: \lim_{x \to a} \: f(x) \: exists \: for \: all \: a≠0}}}$

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