Limits and Derivatives

38. Given, $f(x)=x^n+a{x}^{n-1}+a^2{x}^{n-2}+?+a^{n-1}x+a^n.$

We know that,

$\frac{d}{dx}\left(x^x\right)=n{x}^{x-1}$

So,

$f^{\prime }(x)=\frac{d}{dx}{x}^x+\frac{d}{dx}a{x}^{x-1}+\frac{d}{dx}{a}^2{x}^{x-2}+?+\frac{d}{dx}$ $a^{x-1}x+\frac{d}{dx}{a}^x=n{x}^{n-1}+a(n-1)x^{n-2}+a^2(n-2)x^{n-3}+?+a^{n-1}+0.$

$\begin{array}{l}(?\frac{dax}{dx}=a\frac{dx}{dx}\frac{\mathrm{and\; d}a}{dx}=\mathrm{0whereaisconstant}\end{array}$$\begin{array}{l}\end{array}$$\begin{array}{l}\end{array}$

37. 

Given $f(x)=\frac{x^{100}}{100}+\frac{x^{99}}{99}+?+\frac{x^2}{2}+x+1$

$\underset{h\to 0}{lim}\frac{{(x+h)}^{100}-x^{100}}{100\cdot h}+\underset{h\to 0}{lim}\frac{{(x+h)}^{99}-x^{99}}{99\cdot h}+\dots +\underset{x\to 0}{lim}\frac{{(x+h)}^2-x^2}{2h}$ $+\underset{x\to 0}{lim}x+\underset{x\to 0}{lim}1$

$=\frac{100\cdot x^{99}}{100}+\frac{99x^{98}}{99}+?+\frac{2x}{2}+0+1$

$=x^{99}+x^{98}+?+x+1$

At x=0,

f(X)  =1.

and at x=1,

$f^{\prime }(1)=1^{99}+1^{98}+\dots +1^2+1+1$

=100 $*$ 1

=100 $*f^{\prime }(0)$

Hence, $f^{\prime }(1)=100f^{\prime }(0)$

36. (i) $x^3-27$ (ii) (x -1)(x-2)

(iii) $\frac{1}{x^2}$ (iv) $\frac{x+1}{x-1}$

A.4.(i) Given, $f(x)=x^3-27.$

So, $f^{\prime }(x)=\frac{\underset{h\to 0}{lim}f(x+h)-f(x)}{h}$

$=\underset{h\to 0}{lim}\frac{\left[{(x+h)}^3-27\right]-\left[x^3-27\right]}{h}$

$=\underset{h\to 0}{lim}\frac{x^3+h^3+3xh(x+h)-27-x^3+27}{h}$

$=\underset{h\to 0}{lim}\frac{h\left(h^2+3x(x+h)\right)}{h}$

$=\underset{h\to 0}{lim}{h}^2+3x(x+h)$

=0+3 x(x+ 0)

=3x²

(ii) Given, f(x) =(x-1)(x-2)

=x²- 3x+2

So, $f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$\underset{h\to 0}{lim}\frac{\left[{(x+h)}^2-3(x+h)+2\right]-\left[x^2-3x+2\right]}{h}$

= $\underset{h\to 0}{lim}\frac{x^2+h^2+2xh-3x-3h+2-x^2+3x-2}{h}$

= $\underset{h\to 0}{lim}\frac{h(h+2x-3)}{h}$

$=\underset{h\to 0}{lim}h+2x-3$

= 2x – 3.

(iii) Given, f(x)= $\frac{1}{x^2}$

So, $f^{\prime }(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$=\underset{h\to 0}{lim}\frac{\frac{1}{{(x+h)}^2}-\frac{1}{x^2}}{h}$

$=\frac{\underset{h\to 0}{lim}\frac{x^2-{(x+h)}^2}{{(x+h)}^2-x^2}}{h}$

$=\underset{h\to 0}{lim}\frac{x^2-x^2-h^2-2xh}{h{x}^2(x+h){}^2}$ $$

$=\underset{h\to 0}{lim}\frac{h(-h-2x)}{h{x}^2(x+h){}^2}$

$=\underset{h\to 0}{lim}\frac{-h-2x}{x^2(x+h){}^2}$

$=\frac{-0-2x}{x^2(x+0){}^2}$

$=\frac{-2x}{x^4}$

$=\frac{-2}{x^3}$

(iv) Given, f(x)= $\frac{x+1}{x-1}$

$f(x)=\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{fx+h+1}{x+h-1}-\frac{x+1}{x-1}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{(x+h+1)(x-1)-(x+1)(x+h-1)}{(x+h-1)(x-1)}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{x^2-x+hx-h+x-1-x^2-hx+x-x-h+1}{(x-1)(x+h-1)}\right]$

$=\underset{h\to 0}{lim}\frac{1}{h}\left[\frac{-2h}{(x-1)(x+h-1)}\right]$

$=\underset{h\to 0}{lim}\frac{-2}{(x-1)(x+h-1)}$

$=\frac{-2}{(x-1)(x-1)}=\frac{-2}{{(x-1)}^2}$

35. Given, f (x)= 99x, f   (100)=?

So, f (100)= ${}_{h\to 0}\frac{f(100+h)-f(100)}{h}$

$=\underset{h\to 0}{lim}\frac{99(100+h)-99(100)}{h}$

$=\underset{h\to 0}{lim}\frac{99*100+99*h-99*100}{h}$

$=\underset{h\to 0}{lim}\frac{99h}{h}$

$=\underset{h\to 0}{lim}99$

=99

34. Given, f (x)=x, f   (1)=?

We have,

$f^{\prime }(1)=\underset{h\to 0}{lim}\frac{f(1+h)-f(1)}{h}$

$$ $=\underset{h\to 0}{lim}\frac{1+h-1}{h}$

$=\underset{h\to 0}{lim}\frac{h}{h}$

$=\underset{h\to 0}{lim}1$

=1

33. Given, f (x)=x²- 2 ., $f^{\prime }(10)=?$

We have,  

$f^{\prime }(10)=\underset{h\to 0}{lim}\frac{f(10+h)-f(10)}{h}$

$=\underset{h\to 0}{lim}\frac{\left[{(10+h)}^2-2\right]-\left[10^2-2\right]}{h}$

= $\underset{h\to 0}{lim}\frac{10^2+h^2+20h-2-10^2+2}{h}$

= $\underset{h\to 0}{lim}\frac{h(h+20)}{h}$

$\Rightarrow \underset{h\to 0}{lim}h+20$

=20

32. Given, f (x) = $\begin{array}{l}\{\begin{array}{ll}m{x}^2+n,\hfill & x<0\hfill \\ nx+m,\hfill & 0\le x\le 1\hfill \end{array}\\ n{x}^3+m,x>1.\end{array}$

For $\underset{x\to 0}{lim}f(x)\underset{x\to 0^{-}}{lim,}f(x)=\underset{x\to 0^{+}}{lim}f(x)$

$\Rightarrow \underset{x\to 0^{-}}{lim}\left(m{x}^2+n\right)=\underset{x\to 0^{+}}{lim}\left(m{x}^3+m\right)$

n = m

So, $\underset{x\to 0}{lim}f(x)$ exist for n = m.

Again, $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}nx+m=n+m.$

$\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}n{x}^3+m=n+m$

So, $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{-}}{lim}f(x)=n+m,$ Thus, $\underset{x\to 1}{lim}f(x)$ exist for any integral value of m and n.

31. Given, $\underset{x\to 1}{lim}\frac{f(x)-2}{x^2-1}=\pi$

$\frac{\underset{x\to 1}{lim}f(x)-2}{\underset{x\to 1}{lim}{x}^2-1}=\pi$

$\Rightarrow \underset{x\to 1^{}}{lim}[f(x)-2]=\pi \underset{x\to 1}{lim}\left(x^2-1\right)$

$\Rightarrow \underset{x\to 1}{lim}f(x)-\underset{x\to 1}{lim}2=\pi \left(1^2-1\right)$

$\Rightarrow \underset{x\to 1}{lim}f(x)-2=0.$

$\Rightarrow \underset{x\to 1}{lim}f(x)=2$

30. Given, $f(x)=\{\begin{array}{ccc}\hfill \left|x\right|+1,\hfill & \hfill x<0\hfill & \hfill 0,\hfill \\ \hfill x=0\hfill & \hfill \left|x\right|-1,\hfill & \hfill x>0\hfill \end{array}\underset{x\to a}{lim}f(x)=?$

As | x | = $\{\begin{array}{l}-x,x<0\\ x,x>0\end{array}$

We can rewrite f (x) = $\begin{array}{l}\{\begin{array}{l}-x+1,x<0\\ 0,x=0\end{array}\\ x-1,x>0.\end{array}$

Case 1: when a<0,

$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}(-x+1)=-a+1$

So, $\underset{x\to a}{lim}f(x)$ = exist such that a< 0

Case II when a> 0,

$\underset{x\to a}{lim}f(x)=\underset{x\to a}{lim}(x-1)=a-1$

So, $\underset{x\to a}{lim}f(x)$ exist such that a>0.

Case III when a = 0.

L.H.L = $\underset{x\to a^{-}}{lim}f(x)=\underset{x\to a^{-}}{lim}(-x+1)=\underset{x\to 0^{-}}{lim}(-x+1)=1$

R.H.L = $\underset{x\to a^{+}}{lim}f(x)=\underset{x\to a^{+}}{lim}(x-1)=\underset{x\to 0^{+}}{lim}(x-1)=-1$

Thus, $\underset{x\to a^{-}}{lim}f(x)\ne \underset{x\to a^{+}}{lim}f(x)$

So, $\underset{x\to a}{lim}f(x)$ does not exist at a = 0.

28. Given, f (x) = $\begin{array}{l}\{\begin{array}{l}a+bx,x<1\\ 4,x=1\end{array}\\ b-ax,x>1\end{array}$

Since we need $\underset{x\to 1}{lim}f(x)$ we need,

LHL =

$\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}(a+bx)$ = a + b * 1 = a + b

and RHL =

$\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}(b-ax)$ = b - a * 1 = b - a

Given,  $\underset{x\to 1}{lim}f(x)=f(1):$ we have the following equations

a + b = 4 ____ (1)

b - a = 4 ____ (2)

Adding (1) and (2) we get,

2b = 8

b = 4

Putting b = 4 in (1) we get

a + 4 = 4

a = 0