(i) f(-5^-)
For x < - 5, |x + 5| = - (x + 5)
Thus  f(-5^-) = \(lim_{x\rightarrow-5^- {-(x+5)\over (x+5)}}=-1\)
(ii) f(-5⁺)
For x > - 5, |x + 5| = (x + 5)
Thus  f(-5⁺) = \(lim_{x\rightarrow-5^+{(x+5)\over (x+5)}}=1\)
Note that f( -5^-)  ≠ f( -5⁺). Hence the limit does not exist.

\(f(x)= \begin{cases}\sin x, & x<0 \\ 1-\cos x, & 0 \leq x \leq \pi \\ \cos x, & x>\pi\end{cases}\)
To sketch the graph of y = f(x)
(i) Draw y = sin x where x < 0
(ii) To draw y = \(1-\cos x, 0 \leq x \leq \pi\)
Draw \(y=\cos x, 0 \leq x \leq \pi\)and reflect it on x-axis and then lift up by 1 unit.
(OR)
Putting x = 0, y = 1 - cos 0 = 1 -1 = 0
Putting \(x=\pi / 2, y=1-\cos \pi / 2=1-0=1\)
Putting \(x=\pi, y=1-\cos \pi=1+1=2\)
\(\therefore y=1-\cos x\) passes through \((0,0),\left(\frac{\pi}{2}, 1\right)\), \((\pi, 2)\)
(iii) Draw y = cos X where x >\(\pi\)
when \(x=\pi, y=\cos \pi=-1\)

\( \lim _{x \rightarrow \pi^{-}} f(x)=2 \)
\( \lim _{x \rightarrow \pi^{+}} f(x)=-1 \)
\( \therefore \lim _{x \rightarrow \pi} f(x) \) dose not exist. 
\(\therefore \lim _{x \rightarrow x_0} f(x) \text { exists for } x \in R-\{\pi\} \)

\(lim_{x\rightarrow 0}{\sqrt{1+sin x}-\sqrt{1-sinx}\over tanx}\)
Multiplying the numerator and denominator by \(\sqrt{1+sin x}+\sqrt{1-sin x}\) we get,
\(lim_{x\rightarrow 0}{\sqrt{1+sin x}-\sqrt{1-sinx}\over tanx}\times {\sqrt{1+sin x}+\sqrt{1-sin x}\over \sqrt{1+sin x}+\sqrt{1-sin x}}\)
\(=lim_{x\rightarrow 0}{(1+sin x)-(1-sin x)\over tanx[\sqrt{1+sin x}-\sqrt{1-sinx}]}\)
\(=lim_{x\rightarrow 0}[{2sin x\over{{sin x\over cosx}[ \sqrt{1+sin x}+\sqrt{1-sinx}]}}]\)
\(=2lim_{x\rightarrow 0}{cos x\over \sqrt{1+sin x}+\sqrt{1-sinx}}={2(1)\over \sqrt{1}+\sqrt{1}}={2\over2}=1\)
\(\therefore lim_{x\rightarrow 0}{\sqrt{1+sin x}-\sqrt{1-sinx}\over tanx}=1\)

Given \(y_1=\frac{\sin ^{-1} x}{\sqrt{1-x^2}} .....(1)\)
\(\sqrt{1-x^2} y=\left(\sin ^{-1} x\right)\)
Squaring on both sides, (1 - x²) y² = (sin-1 x)²
Differentiate W.R.T x 
\(\left(1-x^2\right)\left(2 y y_1\right)+y^2(-2 x)=2 \sin ^{-1} x \frac{1}{\sqrt{1-x^2}}\) (using (1))
\(\left(1-x^2\right)\left(2 y y_1\right)-2 x y^2=2 y\)
The above equation divided by 2y.
(1 - x²) y₁ - ay = 1.
Diffrentiate W. R. To x
\( \left(1-x^2\right) y_2+y_1(-2 x)-x y_1-y(1)=0 \)
\(\left(1-x^2\right) y_2-2 x y_1-x y_1-y=0 \)
\(\left(1-x^2\right) y_2-3 x y_1-y=0\)
Hence proved.

Given y = (cos^-1x)²
Differentiating with respect to 'x' we get
y' = 2.cos^-1x\(\left(\frac{-1}{\sqrt{1-x^2}}\right)\)
\(\sqrt{1-x^2} y_1=-\left(2 \cos ^{-1} x\right)\)
Squaring on both sides
\( \left(1-x^2\right) y_1^2=4\left(\cos ^{-1} x\right)^2 \)
\(\left(1-x^2\right) \dot{y}_1^2=4 y \) (using(1))
Differentiate W. R. T x
\( \left(1-x^2\right)\left(2 y_1 y_2\right)+y_1^2(-2 x)=4 y_1 \)
\(\left(1-x^2\right) 2 y_1 y_2-2 x y_1^2-4 y_1=0\)
Divide it by 2y₁
\(\left(1-x^2\right) y_2-x y_1-2=0\)
when x = 0
\( (1-0) y_2-0 y_1-2=0 \)
\(y_2-2=0 \)
\(y_2=2 \)

\(lim_{x\rightarrow5}{\sqrt{x-1}-2\over x-5}\)
Multiplying and dividing by\((\sqrt{x-1}+2)\) we get,
\(lim_{x\rightarrow5}{\sqrt{x-1}-2\over x-5}\times {\sqrt{x-1}+2\over \sqrt{x-1}+2}\)\(=lim_{x\rightarrow5}{(x-1)-4\over x-5[\sqrt{x-1}+2]}\)

\(=lim_{x\rightarrow5}{1\over\sqrt{x-1}+2}={1\over\sqrt{5-1}+2}\)
\({1\over \sqrt{4}+2}={1\over2+2}={1\over 4}\)
\(lim_{x\rightarrow5}{\sqrt{x-1}-2\over x-5}={1\over4}\)

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

Step (i) f(x₀) = 7x₀ + 5.
For any \(\triangle x\neq 0,\),
f(x₀ +\(\triangle\)x) = 7(x₀ + \(\triangle\)x) + 5
= 7x₀ + 7\(\triangle\)x + 5
Step (ii) \(\triangle\)y = f(x₀ + \(\triangle\)x) - f(x₀)
= (7x_o+7\(\triangle\)x_o+5)
Step (iii)\({\triangle \ y\over \triangle x}=7\)
Thus, at any point on the graph of f(x) = 7x + 5, we have
Step (iv) m_tan = \(lim_{\triangle x \rightarrow 0}{\triangle y\over \triangle x}\)
= \(lim_{\triangle x\rightarrow o}(7)\)
= 7.

\(y=\sqrt{1+2 \tan x}\)
\(u =1+2 \tan x \)
\(\frac{d u}{d x} =2 \sec ^2 \cdot x \)
\(y =\sqrt{u}=u^{1 / 2}\)
\(\frac{d y}{d x} =\frac{d y}{d u} \cdot \frac{d u}{d x}=1 / 2 u^{1 / 2-1}\left(2 \sec ^2 x\right) \)
\(=1 / 2 u^{-1 / 2}\left(2 \sec ^2 x\right)=\frac{1}{2 \sqrt{u}}\left(2 \sec ^2 x\right)\)
\(=\frac{\sec ^2 x}{\sqrt{1+2 \tan x}}\)

Differentiating the function implicitly with respect to x, we get
\({dy\over dx}={{dy\over dt}\over {dx\over dt}}={a \ cos \ t\over -a \ sin \ t}=-{cos \ t\over sin \ t}\)
\({d^2y\over dx^2}={d\over dx}({dy\over dx})\)
\(={d\over dx}({-cos \ t \over sin \ t})\)
\(=\frac{d}{d t}\left(\frac{-\cos t}{\sin t}\right) \frac{d t}{d x}=-\left[-\operatorname{cosec}^2 t\right] \times \frac{1}{x^{\prime}(t)}\)
\(=cosec^2t\times {1\over -a \ sin \ t}\)
\(=-\frac{\operatorname{cosec} e^3 t}{a}\)