NCERT Solutions for Class 11 Maths Chapter 12 Exercise 12.1 - Limits and Derivatives

Question 1: Evaluate the following limits

$\lim\limits_{x\rightarrow 3} x +3$

Answer:

$\lim\limits_{x\rightarrow 3} x +3$

$\Rightarrow \lim\limits_{x\rightarrow 3} 3 +3$

$\Rightarrow 6$ Answer

Question 2: Evaluate the following limits $\lim\limits_{x \rightarrow \pi } \left ( x - 22/7 \right )$

Answer:

Below you can find the solution:

$\lim\limits_{x \rightarrow \pi } \left ( x - 22/7 \right )=\pi-\frac{22}{7}$

Question:3 Evaluate the following limits $\lim\limits_{r \rightarrow 1} \pi r^2$

Answer:

The limit

$\lim\limits_{r \rightarrow 1} \pi r^2=\pi(1)^2=\pi$

Answer is $\pi$

Question 4: Evaluate the following limits $\lim\limits_{x \rightarrow {4}} \frac{4x+3 }{x-2}$

Answer:

The limit

$\lim\limits_{x \rightarrow {4}} \frac{4x+3 }{x-2}$

$\Rightarrow \frac{4(4)+3 }{(4)-2}$

$\Rightarrow \frac{19 }{2}$ (Answer)

Question:5 Evaluate the following limits $\lim\limits_{x \rightarrow {-1}}\frac{x^{10}+ x^5 + 1}{x-1}$

Answer:

The limit

$\lim\limits_{x \rightarrow {4}} \frac{x^{10}+ x^5 + 1 }{x-1}$

$\Rightarrow \frac{(-1)^{10}+ (-1)^5 + 1 }{(-1)-1}$

$\Rightarrow \frac{1-1+1}{-2}$

$\Rightarrow -\frac{1}{2}$ (Answer)

Question 6: Evaluate the following limits $\lim\limits_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }$

Answer:

The limit

$\lim\limits_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }$

Lets put

$x+1=y$

since we have changed the function, its limit will also change,

so

$x\rightarrow 0,y\rightarrow 0+1=1$

So our function have became

$\lim\limits_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }$

Now As we know the property

$\lim\limits_{x \rightarrow 1 }\frac{ x^5 -a^n}{x-a }=na^{n-1}$

$\lim\limits_{y \rightarrow 1 }\frac{ y^5 -1}{y-1 }=5(1)^5=5$

Hence,

$\lim\limits_{x \rightarrow 0 }\frac{( x+1)^5 -1}{x }=5$

Question 7: Evaluate the following limits $\lim\limits_{x \rightarrow 2} \frac{3 x^2 - x -10 }{x^2 -4}$

Answer:

The limit

$\lim\limits_{x \rightarrow 2} \frac{3 x^2 - x -10 }{x^2 -4}$

$\Rightarrow \lim\limits_{x \rightarrow 2} \frac{(x-2)(3x+5) }{(x-2)(x+2)}$

$\Rightarrow \lim\limits_{x \rightarrow 2} \frac{(3x+5) }{(x+2)}$

$\Rightarrow \frac{(3(2)+5) }{((2)+2)}$

$\Rightarrow \frac{11 }{4}$ (Answer)

Question 8: Evaluate the following limits $\lim\limits_{x \rightarrow 3} \frac{x ^4 -81}{2x^2 -5x -3}$

Answer:

The limit

$\lim\limits_{x \rightarrow 3} \frac{x ^4 -81}{2x^2 -5x -3}$

At x = 2 both numerator and denominator becomes zero, so lets factorise the function

$\lim\limits_{x \rightarrow 3} \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}$

$\lim\limits_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}$

Now we can put the limit directly, so

$\lim\limits_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{(2x+1)}$

$\Rightarrow \frac{((3)+3)((3)^2+9)}{(2(3)+1)}$

$\Rightarrow \frac{6\times18}{7}$

$\Rightarrow \frac{108}{7}$

Question 9: Evaluate the following limits $\lim\limits_{x \rightarrow 0 } \frac{ax +b}{cx+1}$

Answer:

The limit,

$\lim\limits_{x \rightarrow 0 } \frac{ax +b}{cx+1}$

$\Rightarrow \frac{a(0) +b}{c(0)+1}$

$\Rightarrow b$ (Answer)

Question:10 Evaluate the following limits $\lim\limits_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

Answer:

The limit

$\lim\limits_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}$

Here on directly putting limit , both numerator and the deniminator becomes zero so we factorize the function and then put the limit.

$\lim\limits_{z\rightarrow 1} \frac{z^{1/3}-1}{z^{1/6}-1}=\lim\limits_{z\rightarrow 1} \frac{z^{(1/6)^2}-1^2}{z^{1/6}-1}$

$=\lim\limits_{z\rightarrow 1} \frac{(z^{(1/6)}-1)(z^{(1/6)}+1)}{z^{1/6}-1}$

$\begin{aligned} & =\lim\limits _{z \rightarrow 1}\left(z^{1 / 6}+1\right) \\ & =\left(1^{1 / 6}+1\right)\end{aligned}$

$=1+1$

$=2$ (Answer)

Question 11: Evaluate the following limits $\lim\limits_{x \rightarrow 1} \frac{ax ^2 +bx + c }{cx^2 + bx + a }, a+b +c \neq 0$

Answer:

The limit:

$\lim\limits_{x \rightarrow 1} \frac{ax ^2 +bx + c }{cx^2 + bx + a }, a+b +c \neq 0$

Since Denominator is not zero on directly putting the limit, we can directly put the limits, so,

$\lim\limits_{x \rightarrow 1} \frac{ax ^2 +bx + c }{cx^2 + bx + a }, a+b +c \neq 0$

$=\frac{a(1) ^2 +b(1) + c }{c(1)^2 + b(1) + a },$

$=\frac{a+b+c }{a+b+c },$

$=1$ (Answer)

Question 12: Evaluate the following limits $\lim\limits_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{1}{2}}{x+2}$

Answer:

$\lim\limits_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{1}{2}}{x+2}$

Here, since denominator becomes zero on putting the limit directly, so we first simplify the function and then put the limit,

$\lim\limits_{x\rightarrow -2} \frac{\frac{1}{x}+ \frac{1}{2}}{x+2}$

$=\lim\limits_{x\rightarrow -2} \frac{\frac{x+2}{2x}}{x+2}$

$=\lim\limits_{x\rightarrow -2} \frac{1}{2x}$

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

$= -\frac{1}{4}$ (Answer)

Question 13: Evaluate the following limits $\lim\limits_{x \rightarrow 0 } \frac{\sin ax }{bx }$

Answer:

The limit

$\lim\limits_{x \rightarrow 0 } \frac{\sin ax }{bx }$

Here on directly putting the limits, the function becomes $\frac{0}{0}$ form. so we try to make the function in the form of $\frac{sinx}{x}$ . so,

$\lim\limits_{x \rightarrow 0 } \frac{\sin ax }{bx }$

$=\lim\limits_{x \rightarrow 0 } \frac{\sin ax(ax) }{bx(ax) }$

$=\lim\limits_{x \rightarrow 0 } \frac{\sin ax }{ax }\frac{a}{b}$

As $\lim\limits_{x\rightarrow 0}\frac{sinx}{x}=1$

$=1\cdot\frac{a}{b}$

$=\frac{a}{b}$ (Answer)

Question 14: Evaluate the following limits $\lim\limits_{x \rightarrow 0} \frac{\sin ax }{\sin bx } , a,b \neq 0$

Answer:

The limit,

$\lim\limits_{x \rightarrow 0} \frac{\sin ax }{\sin bx } , a,b \neq 0$

On putting the limit directly, the function takes the zero by zero form So,we convert it in the form of $\frac{sina}{a}$ .and then put the limit,

$\Rightarrow \lim\limits_{x\rightarrow {0}}\frac{\frac{sinax}{ax}}{\frac{sinbx}{bx}}\cdot\frac{ax}{bx}$

$=\frac{\lim\limits_{ax\rightarrow {0}}\frac{sinax}{ax}}{\lim\limits_{bx\rightarrow {0}}\frac{sinbx}{bx}}\cdot\frac{a}{b}$

$=\frac{a}{b}$ (Answer)

Question 15: Evaluate the following limits $\lim\limits_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{\pi ( \pi -x)}$

Answer:

The limit

$\lim\limits_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{\pi ( \pi -x)}$

$\Rightarrow \lim\limits_{x \rightarrow \pi } \frac{\sin ( \pi -x )}{ ( \pi -x)}\times\frac{1}{\pi}$

$= 1\times\frac{1}{\pi}$

$= \frac{1}{\pi}$ (Answer)

Question 16: Evaluate the following limits $\lim\limits_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

Answer:

The limit

$\lim\limits_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

the function behaves well on directly putting the limit,so we put the limit directly. So.

$\lim\limits_{x\rightarrow 0}\frac{\cos x }{\pi -x }$

$=\frac{\cos (0) }{\pi -(0) }$

$=\frac{1 }{\pi }$ (Answer)

Question 17: Evaluate the following limits $\lim\limits_{x\rightarrow 0}$ $\frac{\cos 2x -1}{\cos x -1}$

Answer:

The limit:

$\lim\limits_{x\rightarrow 0}$ $\frac{\cos 2x -1}{\cos x -1}$

The function takes the zero by zero form when the limit is put directly, so we simplify the function and then put the limit

$\lim\limits_{x\rightarrow 0}$ $\frac{\cos 2x -1}{\cos x -1}$

$\lim\limits_{x\rightarrow 0}$ $\frac{-2(sin^2x)}{-2(sin^2(\frac{x}{2}))}$

$=\lim\limits_{x\rightarrow 0}$ $\frac{\frac{(sin^2x)}{x^2}}{\frac{(sin^2(\frac{x}{2}))}{(\frac{x}{2})^2}}\times\frac{x^2}{(\frac{x}{2})^2}$

$=\frac{1^2}{1^2}\times 4$

$= 4$ (Answer)

Question 18: Evaluate the following limits $\lim\limits_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }$

Answer:

$\lim\limits_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }$

The function takes the form zero by zero when we put the limit directly in the function,. since function consist of sin function and cos function, we try to make the function in the form of $\frac{sinx}{x}$ as we know that it tends to 1 when x tends to 0.

So,

$\lim\limits_{x \rightarrow 0}\frac{ax + x \cos x }{b \sin x }$

$=\frac{1}{b}\lim\limits_{x \rightarrow 0}\frac{x(a+ \cos x) }{ \sin x }$

$=\frac{1}{b}\lim\limits_{x \rightarrow 0}\frac{x }{ \sin x }\times(a+ \cos x)$

$=\frac{1}{b}\times1\times(a+ \cos (0))$

$=\frac{a+1}{b}$ (Answer)

Question 19: Evaluate the following limits $\lim\limits_{x \rightarrow 0} x \sec x$

Answer:

$\lim\limits_{x \rightarrow 0} x \sec x$

As function doesn't create any abnormality on putting the limit directly,we can put limit directly. So,

$\lim\limits_{x \rightarrow 0} x \sec x$

$=(0) \sec (0)$

$=(0)\times 1$

$=0$ . (Answer)

Question 20: Evaluate the following limits $\lim\limits_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0$

Answer:

$\lim\limits_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0$

The function takes the zero by zero form when we put the limit into the function directly, so we try to eliminate this case by simplifying the function. So

$\lim\limits_{x\rightarrow 0} \frac{\sin ax + bx }{ax + \sin bx } a,b ,a + b \neq 0$

$=\lim\limits_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot ax+ bx }{ax + \frac{\sin bx}{bx}\cdot bx }$

$=\lim\limits_{x\rightarrow 0} \frac{\frac{\sin ax}{ax} \cdot a+ b }{a + \frac{\sin bx}{bx}\cdot b }$

$=\frac{1\cdot a+b}{a+1\cdot b}$

$=\frac{a+b}{a+ b}$

$=1$ (Answer)

Question 21: Evaluate the following limits $\lim\limits_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

Answer:

$\lim\limits_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

On putting the limit directly the function takes infinity by infinity form, So we simplify the function and then put the limit

$\lim\limits_{x \rightarrow 0} \left ( \csc x - \cot x \right )$

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

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

$=\lim\limits_{x \rightarrow 0} \left (\frac{2sin^2(\frac{x}{2})}{sinx}\right )$

$=\lim\limits_{x \rightarrow 0} \left (\frac{2sin^2(\frac{x}{2})}{(\frac{x}{2})^2}\right )\left ( \frac{(\frac{x}{2})^2}{sinx} \right )$

$=\lim\limits_{x \rightarrow 0} \frac{2}{4}\left (\frac{sin^2(\frac{x}{2})}{(\frac{x}{2})^2}\right )\left ( \frac{(x)}{sinx} \right )\cdot x$

$=\frac{2}{4}\times (1)^2\times0$

$=0$ (Answer)

Question 22: Evaluate the following limits $\lim\limits_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }$

Answer:

$\lim\limits_{x \rightarrow \pi /2 } \frac{\tan 2x }{x - \pi /2 }$

The function takes zero by zero form when the limit is put directly, so we simplify the function and then put the limits,

So

Let's put

$y=x-\frac{\pi}{2}$

Since we are changing the variable, limit will also change.

as

$x\rightarrow \frac{\pi}{2},y=x-\frac{\pi}{2}\rightarrow \frac{\pi}{2}-\frac{\pi}{2}=0$

So function in new variable becomes,

$\lim\limits_{y \rightarrow 0 } \frac{\tan 2(y+\frac{\pi}{2}) }{y+\pi/2- \pi /2 }$

$=\lim\limits_{y \rightarrow 0 } \frac{\tan (2y+\pi) }{y }$

As we know tha property $tan(\pi+x)=tanx$

$=\lim\limits_{y \rightarrow 0 } \frac{\tan (2y) }{y }$

$=\lim\limits_{y \rightarrow 0 } \frac{sin2y}{2y}\cdot\frac{2}{cos2y}$

$=1\times 2$

$=2$ (Answer)

Question 23: Find $\lim\limits_{x \rightarrow 0} f (x ) \: \: \lim\limits_{x \rightarrow 1} f (x) \: \: where \: \: \: f (x) = \left\{\begin{matrix} 2x+3 & x \leq 0 \\ 3 ( x+1)& x > 0 \end{matrix}\right.$

Answer:

Given Function

$f (x) = \left\{\begin{matrix} 2x+3 & x \leq 0 \\ 3 ( x+1)& x > 0 \end{matrix}\right.$

Now,

Limit at x = 0 :

$at\:x=0^-$

: $\lim\limits_{x\rightarrow{0^-}}f(x)=\lim\limits_{x\rightarrow{0^-}}(2x+3)=2(0)+3=3$

$at\:x=0^+$

$\lim\limits_{x\rightarrow{0^+}}f(x)=\lim\limits_{x\rightarrow{0^+}}3(x+1)=3(0+1)=3$

Hence limit at x = 0 is 3.

Limit at x = 1

$at\:x=1^+$

$\lim\limits_{x\rightarrow{1^+}}f(x)=\lim\limits_{x\rightarrow{1^+}}3(x+1)=3(1+1)=6$

$at\:x=1^-$

$\lim\limits_{x\rightarrow{1^-}}f(x)=\lim\limits_{x\rightarrow{1^-}}3(x+1)=3(1+1)=6$

Hence limit at x = 1 is 6.

Question 24: Find $\lim\limits _{x \rightarrow 1} f(x)$, where $f(x)= \begin{cases}x^2-1, & x \leq 1 \\ -x^2-1, & x>1\end{cases}$

Answer:

$\lim\limits _{x \rightarrow 1} f(x)$, where $f(x)= \begin{cases}x^2-1, & x \leq 1 \\ -x^2-1, & x>1\end{cases}$

Limit at $x=1^+$

$\lim\limits_{x \rightarrow 1^+} f (x ) = \lim\limits_{x \rightarrow 1} (-x^2-1)=-(1)^2-1=-2$

Limit at $x=1^-$

$\lim\limits_{x \rightarrow 1^-} f (x ) = \lim\limits_{x \rightarrow 1} (x^2-1)=(1)^2-1=0$

As we can see that Limit at $x=1^+$ is not equal to Limit at $x=1^-$ ,The limit of this function at x = 1 does not exists.

Question:25 Evaluate $\lim\limits_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{|x|}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

Answer:

$\lim\limits_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{|x|}{x} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

The right-hand Limit or Limit at $x=0^+$

$\lim\limits_{x \rightarrow 0^+} f (x) = \lim\limits_{x \rightarrow 0^+} \frac{|x|}{x}=\frac{x}{x}=1$

The left-hand limit or Limit at $x=0^-$

$\lim\limits_{x \rightarrow 0^-} f (x) = \lim\limits_{x \rightarrow 0^-} \frac{|x|}{x}=\frac{-x}{x}=-1$

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

Question 26: Evaluate $\lim\limits_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{x}{|x|} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

Answer:

$\lim\limits_{x \rightarrow 0} f (x) , \: \: where \: \: f (x) = \left\{\begin{matrix} \frac{x}{|x|} & x \neq 0 \\ 0 & x = 0 \end{matrix}\right.$

The right-hand Limit or Limit at $x=0^+$

$\lim\limits_{x \rightarrow 0^+} f (x) = \lim\limits_{x \rightarrow 0^+} \frac{x}{|x|}=\frac{x}{x}=1$

The left-hand limit or Limit at $x=0^-$

$\lim\limits_{x \rightarrow 0^-} f (x) = \lim\limits_{x \rightarrow 0^-} \frac{x}{|x|}=\frac{x}{-x}=-1$

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

Question:27 Find $\lim\limits_{x \rightarrow 5 } f (x) , where f( x) = |x| -5$

Answer:

$\lim\limits_{x \rightarrow 5 } f (x) , where f( x) = |x| -5$

The right-hand Limit or Limit at $x=5^+$

$\lim\limits_{x \rightarrow 5^+} f (x) = \lim\limits_{x \rightarrow 5^+} |x|-5=5-5=0$

The left-hand limit or Limit at $x=5^-$

$\lim\limits_{x \rightarrow 5^-} f (x) = \lim\limits_{x \rightarrow 5^-}|x|-5=5-5=0$

Since Left-hand limit and right-hand limit are equal, The limit of this function at x = 5 is 0.

Question 28: Suppose $f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.$ f (x) = f (1) what are possible values of a and b?

Answer:

Given,

$f (x) = \left\{\begin{matrix} a+bx & x < 1 \\ 4 & x = 1 \\ b - ax & x > 1 \end{matrix}\right.$

And

$\lim\limits_{x\rightarrow 1} f(x)=f(1)$

Since the limit exists,

left-hand limit = Right-hand limit = f(1).

Left-hand limit = f(1)

$\lim\limits_{x\rightarrow 1^-} f(x)= \lim\limits_{x\rightarrow 1}(a+bx)=a+b(1)=a+b=4$

Right-hand limit

$\lim\limits_{x\rightarrow 1^+} f(x)= \lim\limits_{x\rightarrow 1}(b-ax)=b-a(1)=b-a=4$

From both equations, we get that,

$a=0$ and $b=4$

Hence the possible value of a and b are 0 and 4 respectively.

Question 29: Let a1, a2, . . ., an be fixed real numbers and define a function $f (x) = (x - a_1 ) (x - a_2 )...(x - a_n ) .$ What is $\lim\limits_{x \rightarrow a _ 1 }$ f (x) ? For some $a \neq a _ 1 , a _ 2 .... a _n$ , compute l $\lim_{ x \rightarrow a } f (x)$

Answer:

Given,

$f (x) = (x - a_1 ) (x - a_2 )...(x - a_n ) .$

Now,

$\lim\limits _{x \rightarrow a_1} f(x)=\lim\limits _{x \rightarrow a_1}\left[\left(x-a_1\right)\left(x-a_2\right) \ldots\left(x-a_n\right)\right] .=$$\left[\lim\limits _{x \rightarrow a_1}\left(x-a_1\right)\right]\left[\lim\limits _{x \rightarrow a_1}\left(x-a_2\right)\right]\left[\lim\limits _{x \rightarrow a_1}\left(x-a_n\right)\right]=0$

Hence

, $\lim\limits_{x \rightarrow a _ 1 }f(x)=0$

Now,

$\lim\limits_{ x \rightarrow a } f (x)=\lim\limits_{ x \rightarrow a } (x-a_1)(x-a_2)...(x-a_n)$

$\lim\limits_{ x \rightarrow a } f (x)=(a-a_1)(a-a_2)(a-a_3)$

Hence

$\lim\limits_{ x \rightarrow a } f (x)=(a-a_1)(a-a_2)(a-a_3)$ .

Question 30: If $f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.$ For what value (s) of a does $\lim\limits_{x \rightarrow a } f (x)$ exists ?

Answer:

$f (x) = \left\{\begin{matrix} |x| + 1 & x < 0 \\ 0 & x = 0 \\ |x| -1& x > 0 \end{matrix}\right.$

Limit at x = a exists when the right-hand limit is equal to the left-hand limit. So,

Case 1: when a = 0

The right-hand Limit or Limit at $x=0^+$

$\lim\limits_{x \rightarrow 0^+} f (x) = \lim\limits_{x \rightarrow 0^+} |x|-1=1-1=0$

The left-hand limit or Limit at $x=0^-$

$\lim\limits_{x \rightarrow 0^-} f (x) = \lim\limits_{x \rightarrow 0^-} |x|+1=0+1=1$

Since Left-hand limit and right-hand limit are not equal, The limit of this function at x = 0 does not exists.

Case 2: When a < 0

The right-hand Limit or Limit at $x=a^+$

$\lim\limits_{x \rightarrow a^+} f (x) = \lim\limits_{x \rightarrow a^+} |x|-1=a-1$

The left-hand limit or Limit at $x=a^-$

$\lim\limits_{x \rightarrow a^-} f (x) = \lim\limits_{x \rightarrow a^-} |x|-1=a-1$

Since LHL = RHL, Limit exists at x = a and is equal to a-1.

Case 3: When a > 0

The right-hand Limit or Limit at $x=a^+$

$\lim\limits_{x \rightarrow a^+} f (x) = \lim\limits_{x \rightarrow a^+} |x|+1=a+1$

The left-hand limit or Limit at $x=a^-$

$\lim\limits_{x \rightarrow a^-} f (x) = \lim\limits_{x \rightarrow a^-} |x|+1=a+1$

Since LHL = RHL, Limit exists at x = a and is equal to a+1

Hence,

The limit exists at all points except at x=0.

Question:31 If the function f(x) satisfies $\lim\limits_{x \rightarrow 1} \frac{f (x)-2}{x^2-1} = \pi$ , evaluate $\lim_{x \rightarrow 1} f (x)$

Answer:

Given

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

Now,

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

${\lim\limits_{x \rightarrow 1}(f (x)-2)}=\pi{\lim\limits_{x \rightarrow 1}(x^2-1)}$

${\lim\limits_{x \rightarrow 1}(f (x)-2)}=\pi{(1-1)}$

${\lim\limits_{x \rightarrow 1}(f (x)-2)}=0$

${\lim\limits_{x \rightarrow 1}f (x)}=2$

Question:32 If $f(x)=\left[\begin{array}{cl}m x^2+n & ; x<0 \\ n x+m & ; x \leq 0 \leq 1 t \\ n x^3+m & ; x>1\end{array}\right]$ . For what integers m and n does both $\lim\limits _{x \rightarrow 0} f(x)$ and $\lim\limits _{x \rightarrow 1} f(x)$ exist?

Answer:

Given,

$f(x)=\left[\begin{array}{cl}m x^2+n & ; x<0 \\ n x+m & ; x \leq 0 \leq 1 t \\ n x^3+m & ; x>1\end{array}\right]$

Case 1: Limit at x = 0

The right-hand Limit or Limit at $x=0^+$

$\lim\limits_{x \rightarrow 0^+} f (x) = \lim\limits_{x \rightarrow 0^+} nx+m=n(0)+m=m$

The left-hand limit or Limit at $x=0^-$

$\lim\limits_{x \rightarrow 0^-} f (x) = \lim\limits_{x \rightarrow 0^-} mx^2+n=m(0)^2+n=n$

Hence Limit will exist at x = 0 when m = n .

Case 2: Limit at x = 1

The right-hand Limit or Limit at $x=1^+$

$\lim\limits_{x \rightarrow 1^+} f (x) = \lim\limits_{x \rightarrow 1^+} nx^3+m=n(1)^3+m=n+m$

The left-hand limit or Limit at $x=1^-$

$\lim\limits_{x \rightarrow 1^-} f (x) = \lim\limits_{x \rightarrow 1^-} nx+m=n(1)+m=n+m$

Hence Limit at 1 exists at all integers.