Ncert Solutions Maths class 11th

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

27. $\underset{x\to 5}{lim}f(x)=\underset{x\to 5}{lim}\left|x\right|-5$

= | 5 |$-\mathrm{}$ 5

= 5 $-\mathrm{}$5

= 0

26. Given, f (x) $\{\begin{array}{ll}\frac{x}{\left|x\right|},\hfill & x\ne 0\hfill \\ 0,\hfill & x=0\hfill \end{array}$

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

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

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

i e,  $\underset{x\to 0}{lim}f(x)$ does not exist.

25. Given f (x) = $\{\begin{array}{cc}\hfill \frac{\left|x\right|}{x},x\ne 0\hfill & \hfill 0,x=0\hfill \end{array},\underset{x\to 0}{lim}f(x)=?$

We know that, $\left|x\right|=\{\begin{array}{cc}\hfill x,\hfill & \hfill x\ge 0\hfill \\ \hfill -x,\hfill & \hfill x<0\hfill \end{array}$

Now,

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

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

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

i e, $\underset{x\to 0}{lim}f(x)$ does not exist.

24. Given $f(x)=\{\begin{array}{cc}\hfill x^2-1,x\le 1\hfill & \hfill -x^2-1,x>1\hfill \end{array},\underset{x\to 1}{lim}f(x)=?$

Now, L.H.L = $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}\left(x^2-1\right)$

1²- 1 = 0

And R.H.L = $\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}\left(-x^2-1\right)$  (1)2 1 = 1 = 2.

Thus,  $\underset{x\to 1^{-}}{lim}f(x)\ne \underset{x\to 1^{+}}{lim}f(x)$

So,  $\underset{x\to 1}{lim}f(x)$ does not exist.

23. Given f (x) $\{\begin{array}{ll}2x+3,\hfill & x\le 0\hfill \\ 3(x+1),\hfill & x>0\hfill \end{array}\}$

for $\underset{x\to 0}{lim}f(x),$

left hand limit, L.H.S = $\underset{x\to 0^{}}{lim}f(x)$ = $\underset{x\to 0^{}}{lim}(2x+3)$

= 2 0 + 3 = 3.

Right hand limit, R.H.L = $\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{+}}{lim}3(x+1)$

= (0 + 1) = 3 1 = 3.

Thus, $\underset{x\to 0^{-}}{lim}f(x)=\underset{x\to 0^{+}}{lim}f(x)=\underset{x\to 0^{-}}{lim}f(x)=3$

For $\underset{x\to 1}{lim}f(x),$

L.H.L = $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{-}}{lim}3(x+1)=3(1+1)=3*2=6$

R.H.L = $\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1^{+}}{lim}3(x+1)=3(1+1)=3*2=6.$

Thus, $\underset{x\to 1^{-}}{lim}f(x)=\underset{x\to 1^{+}}{lim}f(x)=\underset{x\to 1}{lim}f(x)=6.$

22. $\underset{x\to \raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{lim}\frac{tan2x}{x-\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

Put y = x $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ . So that as y 0 cos $\raisebox{1ex}{$\pi $}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

21. $\underset{x\to 0}{lim}(cosecx-cotx)=\underset{x\to 0}{lim}\left(\frac{1}{sinx}-\frac{cosx}{sinx}\right)$

$=\underset{x\to 0}{lim}\left(\frac{1-cosx}{sinx}\right)$

$=\underset{x\to 0}{lim}\frac{2si{n^2}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{2si{n}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}co{s}^{\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ $\{\begin{array}{}\end{array}?cos2x=1-2si{n}^{2}x\Rightarrow 1-cos2x=2si{n}^{2}x1-cosx=2si{n}^2\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.?sin2x=2sinxcosxsinx=2sin\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\{$

$=\underset{x\to 0}{lim}\raisebox{1ex}{$x$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$

$=tan\raisebox{1ex}{$0$}\!\left/ \!\raisebox{-1ex}{$2$}\right.=0.$