Given \(f(x)= \begin{cases}(x-1)^{3}, & \text { if } x<0 \\ (x+1)^{3}, & \text { if } x \geq 0\end{cases}\)
\(lim_{x\rightarrow 0^-}f(x)=lim_{x\rightarrow 0^-}(x-1)^3=(0-1)^3=-1\)
\(lim_{x\rightarrow 0^+}f(x)=lim_{x\rightarrow 0^+}(x+1)^3=(0+1)^3=1\)
Also, f(0)=(x+1)³=(0+1)³=1
\(\therefore lim_{x\rightarrow 0^-}f(x)\neq lim_{x\rightarrow 0^+}f(x)=f(0)\)
\(\therefore f(x)\) is not continuous at x=0

x	-1	-2	0	1	2
f(x)	(x-1)3 -8	(x-1)3 -27	(x+1)3 1	(x+1)3 8	(x+1)3 27

Given \(x_{0}=3, f(x)= \begin{cases}\frac{x^{2}-9}{x-3}, & \text { if } x \neq 3 \\ 5, & \text { if } x=3\end{cases}\)
\(lim_{x\rightarrow 3^-}f(x)=lim_{x\rightarrow 3^-}{x^2-9\over x-3}=lim_{x\rightarrow 3^-}{(x+3)(x-3)\over x-3}=lim_{x\rightarrow 3^-}x+3=6\)
\(lim_{x\rightarrow 3^+}f(x)=lim_{x\rightarrow 3^+}{x^2-9\over x-3}=lim_{x\rightarrow 3^+}(x+3)=6\)
But f(3)=5
\(\therefore lim_{x\rightarrow 3^-}f(x)=lim_{x\rightarrow 3^+}f(x)\neq f(3)\)
\(\therefore f(x)\) continuous at x_o=3

\({4\sqrt{2}-(cos \ x+sin \ x)^5\over 1-sin 2x}={2^{5\over2}-[(cos x+sin x)^2]^{5\over2}\over 1-sin 2x}\)
\(={2^{5\over2}-(1+sin 2x)^{5\over2}\over 2-(1+sin 2x)}\)
\(={2^{5\over2}-(1+sin 2x)^{5\over2}\over 2-(1+sin 2x)}\)
Therefore, \(lim_{x \rightarrow {\pi\over 4}}{2^{5\over2}-[(cos x+sin x)^2]^{5/2}\over 2-[1+sin 2x]}\)=\(lim_{x \rightarrow {\pi\over 4}}{2^{5\over2}-[1+sin 2x]^{5\over2}\over 2-(1+sin 2x)}\)
Take y = 1 + sin2x. As \(x\rightarrow{\pi\over 4},y \rightarrow 2\)
\(=lim_{y\rightarrow 2}{2^{5\over2}-y^{5\over2}\over 2-y}\)
\(={5\over2}.2^{{5\over2}-1}={5\over2}\times 2^{3\over2}=5\sqrt{2}\).

\(lim_{x-1}{3\sqrt{7+x^3}-\sqrt{3+x^2}\over x-1}\)\(=lim_{x-1}{({7+x^3})^{1\over3}-8^{1\over3}\over 7+x^3-8} \times{x^3-1\over x-1}-{(3+x^2)^{1\over2}-4^{1\over2}\over 3+x^2-4}\times {x^2-1\over x-1}\)|

[\(\because\) when x\(\rightarrow\) 1, x³ \(\rightarrow\) 1 and 7 + x³\(\rightarrow\) 7+ 1 = 8, also when x \(\rightarrow\) 1, x² \(\rightarrow\) 1 and 3 + x²\(\rightarrow\) 4]\(=lim_{7+x^3\rightarrow8}{({7+x^3})^{1\over3}-8^{1\over3}\over 7+x^3-8} (x^2+x+1)-lim_{3+x^2\rightarrow4}{(3+x^2)^{1\over2}-4^{1\over2}\over 3+x^2-4}\times(x+1)\)
\({1\over3}(8)^{{1\over3}-1}(1^2+1+1)-{1\over2}(4^{{1\over2}-1})(1+1)\) \([\because lim_{x \rightarrow a}x^na^n=na^{n-1}]\)

\(={1\over 8^{2\over3}}-{1\over 4^{1\over2}}={1\over (2^{3})^{2\over3}}-{1\over (2^{2})^{1\over2}}\)\(={1\over2^2}-{1\over2}={1\over4}-{1\over2}\)
\(={1-2\over4}={-1\over4}\)
\(\)

(i) Given \(\overrightarrow { a } =2\hat { i } +3\hat { j } -4\hat { k } \)
\(\overrightarrow { b } =3\hat { i } -4\hat { j } -5\hat { k } \) and
\(\overrightarrow { c } =-3\hat { i } +2\hat { j } +3\hat { k }\)
\(\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =(2\hat { i } +3\hat { j } -4\hat { k } )+(3\hat { i } -4\hat { j } -5\hat { k } )+(-3\hat { i } +2\hat { j } +3\hat { k } )\)
\(=2\hat { i } +\hat { j } -6\hat { k } \)
\(|\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } |=\sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+{ (-6) }^{ 2 } } =\sqrt { 4+1+36 } =\sqrt { 41 } \)
Direction cosines of \(\left( \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } \right) \) is \(\frac { 2 }{ \sqrt { 41 } } ,\frac { 1 }{ \sqrt { 41 } } ,\frac { -6 }{ \sqrt { 41 } } \)
\(\text {(ii) } 3 \vec{a}-2 \vec{b}+5 \vec{c}\)
\(3 \vec{a}-2 \vec{b}+5 \vec{c} =3(2 \hat{i}+3 \hat{j}-4 \hat{k})-2(3 \hat{i}-4 \hat{j}-5 \hat{k})+5(-3 \hat{i}+2 \hat{j}+3 \hat{k})\)
\(=6 \hat{i}+9 \hat{j}-12 \hat{k}-6 \hat{i}+8 \hat{j}+10 \hat{k}-15 \hat{i}+10 \hat{j}+15 \hat{k}\)
\(=-15 \hat{i}+27 \hat{j}+13 \hat{k}\)
\(|3 \vec{a}-2 \vec{b}+5 \vec{c}| =\sqrt{(-15)^2+(27)^2+(13)^2} \)
\(=\sqrt{225+729+169}=\sqrt{1123}\)
\(\text {Direction cosines are }\left(\frac{-15}{\sqrt{1123}}, \frac{27}{\sqrt{1123}}, \frac{13}{\sqrt{1123}}\right)\)

Let the sides of the triangle be \(\hat{a}=\) 2\(\hat{i}\) − \(\hat{j}\) + \(\hat{k}\), \(\hat{b}=\) 3 \(\hat{i}\) − 4\(\hat{j}\) - 4\(\hat{k}\), \(\hat{c}=\) \(\hat{i}\) −3\(\hat{j}\) - 5 \(\hat{k}\) 
\(|\hat{a}|=\sqrt{2^2+(-1)^2+1^2}=\sqrt{4+1+1}=\sqrt{6}\)
\(|\overrightarrow{b}|=\sqrt{3^2+(-4)^2+(-4)^2}=\sqrt{9+16+16}=\sqrt{41}\)
\(|\hat{c}|=\sqrt{1^2+(-3)^2+(-5)^2}=\sqrt{1+9+25}=\sqrt{35}\)
Now \(|\overrightarrow{b}|^2=(\sqrt{41})^2=41=35+6(\sqrt{35})^2+(\sqrt{6})^2=|\hat{a}|^2 +|\hat{c}|^2\)
By Pythagoras theorem, the given vectors from a right angled triangle.

Let the position vector of the vertices of the \(\triangle\)ABC be \(\overrightarrow{a},\overrightarrow{b}\)and \(\overrightarrow{c}\) respectively.
Since D is the mid-point of BC.
\(\Rightarrow \overrightarrow{OD}={\overrightarrow{b}+\overrightarrow{c}\over 2}\)
E is the mid-point of AC,
\(\Rightarrow \overrightarrow{OE}={\overrightarrow{a}+\overrightarrow{c}\over 2}\)
and F is the mid-point of AB
\( \overrightarrow{OF}={\overrightarrow{a}+\overrightarrow{b}\over 2}\)
To prove that \(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{o}\)

LHS=\(\overrightarrow{AD}+\overrightarrow{BE}+\overrightarrow{CF}=\overrightarrow{OD}-\overrightarrow{OA}+\overrightarrow{OE}-\overrightarrow{OB}+\overrightarrow{OF}+\overrightarrow{OC}\)
\(={\overrightarrow{b}+\overrightarrow{c}\over 2}=\overrightarrow{a}+{\overrightarrow{a}+\overrightarrow{c}\over 2}-\overrightarrow{b}\)+\({\overrightarrow{a}+\overrightarrow{b}\over 2}-\overrightarrow{c}\)
\(={\overrightarrow{b}+\overrightarrow{c}-2\overrightarrow{a}+\overrightarrow{a}+\overrightarrow{c}-2\overrightarrow{b}+\overrightarrow{a}+\overrightarrow{b}-2\overrightarrow{c}\over 2}\) \(={\overrightarrow{0}\over2}=\overrightarrow{0}=RHS\)
Hence proved.

Given  \(\overrightarrow{PO}\) +\(\overrightarrow{OQ}\) = \(\overrightarrow{QO}\) +\(\overrightarrow{OR}\)
\(\Rightarrow \overrightarrow{PQ}=\overrightarrow{QR}\) [By triangle law of addition]
\(\Rightarrow \overrightarrow{PQ}=\overrightarrow{QR}\) and Q is a common point.
Hence, the points P, Q, R are collinear.

Let \(\overrightarrow{a}\) and \(\overrightarrow{b}\) be the position vectors of the points A and B.
\(\Rightarrow \overrightarrow{OA}=\overrightarrow{a}\) and \( \overrightarrow{OB}=\overrightarrow{b}\).
Let P divides the line segment AB in the ratio 1:2 and Q divides the line segment AB in the ratio 2 : 1
\(\therefore \overrightarrow{OP}={1.(\overrightarrow{OB})+2(\overrightarrow{OA})\over 1+2}={1(\overrightarrow{b})+2(\overrightarrow{a})\over 3}={\overrightarrow{b}+2\overrightarrow{a}\over 3}\)
and \( \overrightarrow{OQ}={2(\overrightarrow{OB})+1(\overrightarrow{OA})\over 2+1}={2\overrightarrow{b}+\overrightarrow{a}\over 3}={\overrightarrow{a}+2\overrightarrow{b}\over 3}\)
Hence, the required position vectors are \({\overrightarrow{b}+2\overrightarrow{a}\over 3}\)and \({\overrightarrow{a}+2\overrightarrow{b}\over 3}\).

\(lim_{x\rightarrow 0}{2^x-3^x\over x}=lim_{x\rightarrow0}{2^x-1-3^x+1\over x}\) [Adding and subtracting 1 in the numerator]
\(lim_{x\rightarrow 0}{2^x-1\over x}-{(3^x+1)\over x}=lim_{x\rightarrow0}{2^x-1\over x}-lim_{x\rightarrow0}({3^x-1\over x})\)
\(=log2-log3\) \([\because lim_{x\rightarrow 0}{a^x-1\over x}=log \ a]\)
\(=log({2\over3})\)
\(\therefore lim_{x\rightarrow 0}{2^x-3^x\over x}=log({2\over3})\)

One can tabulate values of x near 0 (from either side) and conclude \(f(x)={1\over x^2+x^3}\) grows without bound and hence \(f(x)\rightarrow \infty as \ x\rightarrow 0.\)
To calculate this limit without making a table, we first divide the numerator and denominator by x². This division can be done, since in the calculation of the limit x ≠ 0 and hence x² ≠ 0. We can have
\(lim_{x\rightarrow 0}{1\over x^2+x^3}=lim_{x\rightarrow 0}{{1\over x^2}\over{x^2+x^3\over x^2}}={lim_{x\rightarrow}({1\over x^2})\over lim_{x\rightarrow0}(1+x)}\)
Now \({1\over 8}\rightarrow \infty as \ x\rightarrow 0\) and \(lim_{x\rightarrow0}(1+x)=1\) .
Thus the numerator grows without bound while the denominator approaches 1, implying that \({1\over (x^2+x^3)}\) does tend to infinity.

The function is graphed

Clearly \(lim_{x\rightarrow 0^-}f(x)=-1\) and \(lim_{x\rightarrow 0^+}f(x)=1\) .
Since these limits are different, \(lim_{x\rightarrow 0}f(x)\) does not exist.

The given vector is 3\(\hat{i}\) - 3\(\hat{k}\) + 4\(\hat{j} \Rightarrow\) 3\(\hat{i}\) + 4\(\hat{j}\) - 3\(\hat{k}\)
The direction ratios are 3, 4, -3.
r = \(\sqrt{x^2+y^2+z^2}=\sqrt{3^2+4^2+(-3)^2}\)
\(=\sqrt{9+16+9}=\sqrt{34}\)
Hence, the direction cosines are \({3\over \sqrt{34}},{4\over \sqrt{34}},{-3\over \sqrt{34}}\)

The left-hand limit and right-hand limit does not coincide at x=x_o.

The limit of f(x) does not exist at x = x_o.

\(lim_{x\rightarrow1}{x^m-1\over x^n-1}\)
Multiplying and dividing by (x - 1) we get,
\(lim_{x\rightarrow1}{x^m-1^m\over x-1} \times { x-1\over x^n-1^n}=\)\((lim_{x\rightarrow1}{x^m-1^m\over x-1}) \times lim_{x\rightarrow1} {1\over ({x^n-1^n\over x-1})}\)
\(=m.(1)^{m-1}\times {1\over n(1)^{n-1}}={m \over n}\) \([\because lim_{x\rightarrow a}{x^n-a^n\over x-a}=n.a^{n-1}]\)

\(lim_{x\rightarrow{0}}sec \ x\)
At x = 0, the curve meets the y-axis at 1.
\(\therefore lim_{x\rightarrow{0}}sec \ x=1\)

\(lim_{x\rightarrow1}(x^2+2)\)
Atx = 1, the value of the curve on y-axis is 3.
\(\therefore lim_{x\rightarrow1}(x^2+2)=3\)

\(|x|= \begin{cases}-x & \text { if } x<0 \\ 0 & \text { if } x=0 \\ x & \text { if } x>0\end{cases}\)
If x > 0,then |x| =  x, which tends to 0 as
\(x \rightarrow 0\) from the right of 0. That is, \(\lim _{ x\rightarrow0^+}{|x| } =0\)
If x < 0, then |x| = - x which again tends to 0 as x\(\rightarrow\)0. from the left of 0. That is, \(\lim _{ x\rightarrow0^-}{|x| } =0\).
Thus, \(\lim _{ x\rightarrow0^-}{|x| } =0=\lim _{ x\rightarrow0^+}{|x| }.\)
Hence \(\lim _{ x\rightarrow0}{|x| } =0\).

Given \(\overrightarrow{a}=\hat{i}-2\hat{j}+\hat{k}\)
\(\overrightarrow{b}=3\hat{i}-4\hat{j}-2\hat{k}\)|
\(\overrightarrow{a} . \overrightarrow{b}=(\hat{i}-2\hat{j}+\hat{k}).(3\hat{i}-4\hat{j}-2\hat{k})\)
= 1(3) - 2(-4) + 1(-2)
= 3 + 8 - 2 = 9
\(\therefore \overrightarrow{a} . \overrightarrow{b}=9\)

\(f(-1) =\lim _{x \rightarrow-1} f(x) \)
\(=\lim _{x \rightarrow-1} \frac{x^{2}-1}{x^{3}+1} \)
\(=\lim _{x \rightarrow-1} \frac{x^{2}-(-1)^{2}}{x^{3}-(-1)^{3}}=\frac{2(-1)^{2-1}}{3(-1)^{3-1}}=\frac{-2}{3} \)

\(\lim _{n \rightarrow \infty}[\left.\frac{1}{n^{2}}+\frac{2}{n^{2}}+\frac{3}{n^{2}}+\cdots+\frac{n}{n^{2}}\right] \)
\(=\lim _{n \rightarrow \infty}\left[\frac{(1+2+3+\cdots+\dot{n})}{n^{2}}\right] \)
\(=\lim _{n \rightarrow \infty}\left[\frac{n(n+1)}{2 n^{2}}\right] \)
\(=\frac{1}{2} \lim _{n \rightarrow \infty}\left(\frac{n+1}{n}\right) \)
\(=\frac{1}{2} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=1 / 2(1+0)=1 / 2 \)

\(\lim _{x \rightarrow 0} \frac{8^{x}-4^{x}-2^{x}+1^{x}}{x^{2}}=\lim _{x \rightarrow 0} \frac{4^{x}\left(2^{x}-1\right)-1\left(2^{x}-1\right)}{x^{2}}\)
\(=\lim _{x \rightarrow 0} \frac{\left(2^{x}-1\right)\left(4^{x}-1\right)}{x^{2}}=\lim _{x \rightarrow 0}\left(\frac{2^{x}-1}{x}\right)\left(\frac{4^{x}-1}{x}\right) \)
\(=\log 2 \times \log 4=\log 2 \times 2 \log 2=2(\log 2)^{2} \)

\(\text { Unit vector }=\frac{\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}}{\sqrt{\lambda^{2}+4 \lambda^{2}+4 \lambda^{2}}}\)
\(=\frac{\lambda \hat{k}+2 \lambda \hat{j}+2 \lambda \hat{k}}{\sqrt{9 \lambda^{2}}}=\frac{\lambda \hat{i}+2 \lambda \hat{j}+2 \lambda \hat{k}}{3 \lambda} \)
\(=\frac{\lambda(\hat{i}+2 \hat{j}+2 \hat{k})}{3 \lambda}=\frac{1}{3}(\hat{i}+2 \hat{j}+2 \hat{k}) \)
\(\lambda =\frac{1}{3} \)