Class 11 Maths NCERT Solutions Chapter 12 Ex 12.1 – Limits and Derivatives

Since f(x)=x+3 is a polynomial function, its limit at x=3 is simply its value at x=3:

\lim_{x\to3}(x+3)=3+3=6

Since f(x)=x-\dfrac{22}{7} is a polynomial function, its limit at x=\pi is its value at x=\pi:

\lim_{x\to\pi}\left(x-\dfrac{22}{7}\right)=\pi-\dfrac{22}{7}

Since f(r)=\pi r^2 is a polynomial function in r, its limit at r=1 is its value at r=1:

\lim_{r\to1}\pi r^2=\pi(1)^2=\pi

This is a rational function, and the denominator does not vanish at x=4 since 4-2=2\ne0. So the limit equals the value of the function at x=4:

\lim_{x\to4}\dfrac{4x+3}{x-2}=\dfrac{4(4)+3}{4-2}=\dfrac{19}{2}

This is a rational function, and the denominator does not vanish at x=-1 since -1-1=-2\ne0. So the limit equals the value of the function at x=-1:

\lim_{x\to-1}\dfrac{x^{10}+x^5+1}{x-1}=\dfrac{(-1)^{10}+(-1)^5+1}{-1-1}=\dfrac{1-1+1}{-2}=\dfrac{1}{-2}

Substituting x=0 gives the indeterminate form \dfrac{0}{0}, so the expression must be rewritten using the standard result \lim_{x\to a}\dfrac{x^n-a^n}{x-a}=na^{n-1}.

Put y=x+1, so that y\to1 as x\to0, and x=y-1:

\lim_{x\to0}\dfrac{(x+1)^5-1}{x}=\lim_{y\to1}\dfrac{y^5-1^5}{y-1}=5(1)^{5-1}=5

Substituting x=2 gives \dfrac{12-2-10}{4-4}=\dfrac{0}{0}, so both the numerator and denominator share the factor (x-2).

3x^2-x-10=(x-2)(3x+5) and x^2-4=(x-2)(x+2)

\lim_{x\to2}\dfrac{3x^2-x-10}{x^2-4}=\lim_{x\to2}\dfrac{(x-2)(3x+5)}{(x-2)(x+2)}=\lim_{x\to2}\dfrac{3x+5}{x+2} (as x\ne2)

=\dfrac{3(2)+5}{2+2}=\dfrac{11}{4}

Substituting x=3 gives \dfrac{81-81}{18-15-3}=\dfrac{0}{0}, so both the numerator and denominator share the factor (x-3).

x^4-81=(x^2-9)(x^2+9)=(x-3)(x+3)(x^2+9) and 2x^2-5x-3=(x-3)(2x+1)

\lim_{x\to3}\dfrac{x^4-81}{2x^2-5x-3}=\lim_{x\to3}\dfrac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)}=\lim_{x\to3}\dfrac{(x+3)(x^2+9)}{2x+1} (as x\ne3)

=\dfrac{(3+3)(9+9)}{2(3)+1}=\dfrac{6\times18}{7}=\dfrac{108}{7}

The denominator does not vanish at x=0 since c(0)+1=1\ne0. So the limit equals the value of the function at x=0:

\lim_{x\to0}\dfrac{ax+b}{cx+1}=\dfrac{a(0)+b}{c(0)+1}=\dfrac{b}{1}=b

Substituting z=1 gives the indeterminate form \dfrac{0}{0}. Divide numerator and denominator by (z-1) and use \lim_{z\to1}\dfrac{z^n-1}{z-1}=n (valid for any rational n):

\lim_{z\to1}\dfrac{z^{1/3}-1}{z^{1/6}-1}=\lim_{z\to1}\left[\dfrac{z^{1/3}-1}{z-1}\div\dfrac{z^{1/6}-1}{z-1}\right]=\dfrac{\lim_{z\to1}\dfrac{z^{1/3}-1}{z-1}}{\lim_{z\to1}\dfrac{z^{1/6}-1}{z-1}}

=\dfrac{\frac{1}{3}(1)^{1/3-1}}{\frac{1}{6}(1)^{1/6-1}}=\dfrac{1/3}{1/6}=2

At x=1, both the numerator and denominator become a+b+c, which is non-zero by hypothesis. So the denominator does not vanish, and the limit equals the value of the function at x=1:

\lim_{x\to1}\dfrac{ax^2+bx+c}{cx^2+bx+a}=\dfrac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a}=\dfrac{a+b+c}{a+b+c}=1

Substituting x=-2 gives the indeterminate form \dfrac{0}{0}, so first combine the numerator into a single fraction:

\dfrac{1}{x}+\dfrac{1}{2}=\dfrac{2+x}{2x}

\lim_{x\to-2}\dfrac{\frac{1}{x}+\frac{1}{2}}{x+2}=\lim_{x\to-2}\dfrac{2+x}{2x(x+2)}=\lim_{x\to-2}\dfrac{1}{2x} (as x\ne-2, so x+2\ne0 cancels)

=\dfrac{1}{2(-2)}=-\dfrac{1}{4}

Multiply and divide by a so that the standard limit \lim_{\theta\to0}\dfrac{\sin\theta}{\theta}=1 can be applied with \theta=ax (note ax\to0 as x\to0):

\lim_{x\to0}\dfrac{\sin ax}{bx}=\dfrac{a}{b}\lim_{x\to0}\dfrac{\sin ax}{ax}=\dfrac{a}{b}(1)=\dfrac{a}{b}

Multiply and divide by ax in the numerator and by bx in the denominator:

\lim_{x\to0}\dfrac{\sin ax}{\sin bx}=\lim_{x\to0}\left[\dfrac{\sin ax}{ax}\times\dfrac{bx}{\sin bx}\times\dfrac{a}{b}\right]

=\dfrac{a}{b}\times\lim_{ax\to0}\dfrac{\sin ax}{ax}\times\lim_{bx\to0}\dfrac{bx}{\sin bx}=\dfrac{a}{b}\times1\times1=\dfrac{a}{b}

Put y=\pi-x, so that y\to0 as x\to\pi:

\lim_{x\to\pi}\dfrac{\sin(\pi-x)}{\pi(\pi-x)}=\lim_{y\to0}\dfrac{\sin y}{\pi y}=\dfrac{1}{\pi}\lim_{y\to0}\dfrac{\sin y}{y}=\dfrac{1}{\pi}(1)=\dfrac{1}{\pi}

The denominator does not vanish at x=0 since \pi-0=\pi\ne0. So the limit equals the value of the function at x=0:

\lim_{x\to0}\dfrac{\cos x}{\pi-x}=\dfrac{\cos0}{\pi-0}=\dfrac{1}{\pi}

Use the identity 1-\cos\theta=2\sin^2\left(\dfrac{\theta}{2}\right), applied with \theta=2x and \theta=x:

\cos2x-1=-2\sin^2x and \cos x-1=-2\sin^2\left(\dfrac{x}{2}\right)

\lim_{x\to0}\dfrac{\cos2x-1}{\cos x-1}=\lim_{x\to0}\dfrac{-2\sin^2x}{-2\sin^2\left(\frac{x}{2}\right)}=\lim_{x\to0}\dfrac{\sin^2x}{\sin^2\left(\frac{x}{2}\right)}

Writing \sin x=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right):

=\lim_{x\to0}\dfrac{4\sin^2\left(\frac{x}{2}\right)\cos^2\left(\frac{x}{2}\right)}{\sin^2\left(\frac{x}{2}\right)}=\lim_{x\to0}4\cos^2\left(\dfrac{x}{2}\right)=4(1)^2=4

Factor x from the numerator and divide numerator and denominator by x:

\lim_{x\to0}\dfrac{ax+x\cos x}{b\sin x}=\lim_{x\to0}\dfrac{x(a+\cos x)}{b\sin x}=\dfrac{1}{b}\lim_{x\to0}\left[(a+\cos x)\times\dfrac{x}{\sin x}\right]

=\dfrac{1}{b}\times\lim_{x\to0}(a+\cos x)\times\lim_{x\to0}\dfrac{x}{\sin x}=\dfrac{1}{b}\times(a+1)\times1=\dfrac{a+1}{b}

Write \sec x=\dfrac{1}{\cos x}, and since \cos x does not vanish at x=0, the function is continuous there:

\lim_{x\to0}x\sec x=\lim_{x\to0}\dfrac{x}{\cos x}=\dfrac{0}{\cos0}=\dfrac{0}{1}=0

Divide every term by x, then use \lim_{x\to0}\dfrac{\sin ax}{ax}=1 and \lim_{x\to0}\dfrac{\sin bx}{bx}=1:

\lim_{x\to0}\dfrac{\sin ax+bx}{ax+\sin bx}=\lim_{x\to0}\dfrac{\frac{\sin ax}{x}+b}{a+\frac{\sin bx}{x}}=\lim_{x\to0}\dfrac{a\cdot\frac{\sin ax}{ax}+b}{a+b\cdot\frac{\sin bx}{bx}}

=\dfrac{a(1)+b}{a+b(1)}=\dfrac{a+b}{a+b}=1

Write both terms with a common denominator in terms of \sin x and \cos x:

\operatorname{cosec}x-\cot x=\dfrac{1}{\sin x}-\dfrac{\cos x}{\sin x}=\dfrac{1-\cos x}{\sin x}

Using 1-\cos x=2\sin^2\left(\frac{x}{2}\right) and \sin x=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right):

\dfrac{1-\cos x}{\sin x}=\dfrac{2\sin^2\left(\frac{x}{2}\right)}{2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)}=\tan\left(\dfrac{x}{2}\right)

\lim_{x\to0}(\operatorname{cosec}x-\cot x)=\lim_{x\to0}\tan\left(\dfrac{x}{2}\right)=\tan0=0

Put y=x-\dfrac{\pi}{2}, so that y\to0 as x\to\dfrac{\pi}{2}, and x=y+\dfrac{\pi}{2}:

\tan2x=\tan\left(2y+\pi\right)=\tan2y (since \tan has period \pi)

\lim_{x\to\frac{\pi}{2}}\dfrac{\tan2x}{x-\frac{\pi}{2}}=\lim_{y\to0}\dfrac{\tan2y}{y}=2\lim_{y\to0}\dfrac{\tan2y}{2y}=2(1)=2

Left-hand limit (using f(x)=2x+3 for x\le0):

\lim_{x\to0^-}f(x)=\lim_{x\to0^-}(2x+3)=2(0)+3=3

Right-hand limit (using f(x)=3(x+1) for x>0):

\lim_{x\to0^+}f(x)=\lim_{x\to0^+}3(x+1)=3(0+1)=3

Since the left-hand and right-hand limits are equal, \lim_{x\to0}f(x)=3.

Since 1>0, values of x near 1 (on either side) all use the formula f(x)=3(x+1), so this is simply a polynomial limit:

\lim_{x\to1}f(x)=3(1+1)=6

Left-hand limit (using f(x)=x^2-1 for x\le1):

\lim_{x\to1^-}f(x)=\lim_{x\to1^-}(x^2-1)=1^2-1=0

Right-hand limit (using f(x)=-x^2-1 for x>1):

\lim_{x\to1^+}f(x)=\lim_{x\to1^+}(-x^2-1)=-1^2-1=-2

Since \lim_{x\to1^-}f(x)=0 and \lim_{x\to1^+}f(x)=-2 are not equal, the limit does not exist.

For x<0, |x|=-x, so f(x)=\dfrac{-x}{x}=-1. For x>0, |x|=x, so f(x)=\dfrac{x}{x}=1.

\lim_{x\to0^-}f(x)=-1 and \lim_{x\to0^+}f(x)=1

Since the left-hand and right-hand limits are not equal, the limit does not exist.

For x<0, |x|=-x, so f(x)=\dfrac{x}{-x}=-1. For x>0, |x|=x, so f(x)=\dfrac{x}{x}=1. This is the same function as in Question 25, written the other way up.

\lim_{x\to0^-}f(x)=-1 and \lim_{x\to0^+}f(x)=1

Since the left-hand and right-hand limits are not equal, the limit does not exist.

Since 5>0, values of x near 5 are all positive, so |x|=x throughout a neighbourhood of 5 and the function is simply f(x)=x-5 there:

\lim_{x\to5}f(x)=\lim_{x\to5}(x-5)=5-5=0

Left-hand limit (using f(x)=a+bx for x<1):

\lim_{x\to1^-}f(x)=a+b(1)=a+b

Right-hand limit (using f(x)=b-ax for x>1):

\lim_{x\to1^+}f(x)=b-a(1)=b-a

For \lim_{x\to1}f(x) to exist, the left-hand and right-hand limits must be equal:

a+b=b-a\ \Rightarrow\ 2a=0\ \Rightarrow\ a=0

Also, since f(1)=4, the condition \lim_{x\to1}f(x)=f(1) requires the common limit value to equal 4:

a+b=4 and, with a=0, this gives b=4.

Check: b-a=4-0=4, which agrees with a+b=4. Both conditions are satisfied.

f(x) is a polynomial function (a product of linear factors), so its limit at any point equals its value there.

\lim_{x\to a_1}f(x)=f(a_1)=(a_1-a_1)(a_1-a_2)\cdots(a_1-a_n)=0

since the first factor (a_1-a_1) is zero.

\lim_{x\to a}f(x)=f(a)=(a-a_1)(a-a_2)\cdots(a-a_n)

and since a is different from every a_i, none of these factors is zero, so the value is simply this non-zero product.

Left-hand limit (using f(x)=|x|+1=-x+1 for x<0):

\lim_{x\to0^-}f(x)=-0+1=1

Right-hand limit (using f(x)=|x|-1=x-1 for x>0):

\lim_{x\to0^+}f(x)=0-1=-1

Since 1\ne-1, \lim_{x\to0}f(x) does not exist.

For a<0, every point in a small neighbourhood of a also satisfies x<0, so f equals the single polynomial formula -x+1 throughout, and its limit at a equals its value there.

For a>0, similarly every nearby point satisfies x>0, so f equals x-1 throughout, and its limit at a equals its value there.

So the limit exists for every a\ne0.

Since x^2-1\to0 as x\to1, and the given limit is the finite number \pi, the numerator f(x)-2 must also tend to 0 as x\to1 — otherwise the ratio would blow up rather than approach a finite limit.

Formally, using the algebra of limits (writing f(x)-2 as the product of the given ratio and (x^2-1)):

\lim_{x\to1}\big[f(x)-2\big]=\lim_{x\to1}\left[\dfrac{f(x)-2}{x^2-1}\times(x^2-1)\right]=\lim_{x\to1}\dfrac{f(x)-2}{x^2-1}\times\lim_{x\to1}(x^2-1)

=\pi\times(1^2-1)=\pi\times0=0

Therefore \lim_{x\to1}f(x)-2=0, which gives \lim_{x\to1}f(x)=2.

Left-hand limit (using f(x)=mx^2+n for x<0):

\lim_{x\to0^-}f(x)=m(0)^2+n=n

Right-hand limit (using f(x)=nx+m for 0\le x\le1):

\lim_{x\to0^+}f(x)=n(0)+m=m

For the limit at x=0 to exist, these must be equal: n=m.

Left-hand limit (using f(x)=nx+m for 0\le x\le1):

\lim_{x\to1^-}f(x)=n(1)+m=n+m

Right-hand limit (using f(x)=nx^3+m for x>1):

\lim_{x\to1^+}f(x)=n(1)^3+m=n+m

These are automatically equal for every value of m and n, so the limit at x=1 always exists, regardless of what m and n are.

So the only condition needed for both limits to exist is m=n, and this holds for any integer value of m (and the matching n).