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Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 2
  • Last Updated : 07 Apr, 2021

Question 17:\lim_{x \to 0} \frac{cos \hspace{0.1cm}2x-1}{cos \hspace{0.1cm}x-1}

Solution:

In\lim_{x \to 0} \frac{cos \hspace{0.1cm}2x-1}{cos \hspace{0.1cm}x-1} , as x⇢0

As we know, cos 2θ = 1-2sin2θ

Substituting the values, we get

\lim_{x \to 0} \frac{1-2sin^2x-1}{1-2sin^2(\frac{x}{2})-1}



=\lim_{x \to 0} \frac{sin^2x}{sin^2(\frac{x}{2})}

Put x = 0, we get

\lim_{x \to 0} \frac{sin^2x}{sin^2(\frac{x}{2})} = \frac{0}{0}

As, this limit becomes undefined

Now, let’s multiply and divide the numerator by x2 and denominator by(\frac{x}{2})^2 to make it equivalent to theorem.

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

Hence, we have

\lim_{x \to 0} \frac{\frac{sin^2x \times x^2}{x^2}}{\frac{sin^2(\frac{x}{2}) \times (\frac{x}{2})^2}{(\frac{x}{2})^2}}



=\lim_{x \to 0} \frac{(\frac{sin \hspace{0.1cm}x}{x})^2\times x^2}{(\frac{sin \hspace{0.1cm}(\frac{x}{2})}{\frac{x}{2}})^2\times (\frac{x}{2})^2}

=\frac{(\lim_{x \to 0}\frac{sin \hspace{0.1cm}x}{x})^2\times\lim_{x \to 0} x^2}{(\lim_{x \to 0}\frac{sin \hspace{0.1cm}(\frac{x}{2})}{\frac{x}{2}})^2\times\lim_{x \to 0} (\frac{x}{2})^2}

By using the theorem, we get

=\frac{(1)^2\times\lim_{x \to 0} x^2}{(1)^2\times\lim_{x \to 0} (\frac{x^2}{4})}

=\lim_{x \to 0}\frac{x^2}{(\frac{x^2}{4})}

=\lim_{x \to 0}(4)

= 4

Question 18:\lim_{x \to 0} \frac{ax+xcos \hspace{0.1cm}x}{bsin \hspace{0.1cm}x}

Solution:

In\lim_{x \to 0} \frac{ax+xcos \hspace{0.1cm}x}{bsin \hspace{0.1cm}x} , as x⇢0

Put x = 0, we get

\lim_{x \to 0} \frac{ax+xcos \hspace{0.1cm}x}{bsin \hspace{0.1cm}x} = \frac{0}{0}

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem.

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

Hence, we have

\lim_{x \to 0} \frac{x(a+cos \hspace{0.1cm}x)}{bsin \hspace{0.1cm}x}

=\frac{1}{b}\times \lim_{x \to 0} \frac{x}{sin \hspace{0.1cm}x}\times \lim_{x \to 0} (a+cos \hspace{0.1cm}x)

By using the theorem, we get

=\frac{1}{b}\times 1\times \lim_{x \to 0} (a+cos \hspace{0.1cm}x)

=\frac{1}{b}\times a

Putting x=0, we have

=\frac{a}{b}

Question 19:\lim_{x \to 0} x sec\hspace{0.1cm}x

Solution:

In\lim_{x \to 0} x sec\hspace{0.1cm}x , as x⇢0

Put x = 0, we get

\lim_{x \to 0} x sec\hspace{0.1cm}x = 0 ×1

= 0

Question 20:\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax+bx}{ax+sin \hspace{0.1cm}bx}

Solution:

In\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax+bx}{ax+sin \hspace{0.1cm}bx} , as x⇢0

Put x = 0, we get

\lim_{x \to 0} \frac{sin \hspace{0.1cm}ax+bx}{ax+sin \hspace{0.1cm}bx} = \frac{0}{0}

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem.

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

Hence, we can write the equation as follows:

\lim_{x \to 0} \frac{\frac{sin \hspace{0.1cm}(ax)\times ax}{ax}+bx}{ax+\frac{sin \hspace{0.1cm}(bx)\times bx}{bx}}

=\frac{\lim_{x \to 0}\frac{sin \hspace{0.1cm}(ax)}{ax}\times \lim_{x \to 0}ax+\lim_{x \to 0}bx}{\lim_{x \to 0}ax+\lim_{x \to 0}\frac{sin \hspace{0.1cm}(bx)}{bx}\times\lim_{x \to 0} bx}

By using the theorem, we get

=\frac{1\times \lim_{x \to 0}ax+\lim_{x \to 0}bx}{\lim_{x \to 0}ax+1\times\lim_{x \to 0} bx}

=\frac{\lim_{x \to 0}ax+\lim_{x \to 0}bx}{\lim_{x \to 0}ax+\lim_{x \to 0} bx}

=\lim_{x \to 0} \frac{ax+bx}{ax+bx}

=\lim_{x \to 0} 1

Putting x=0, we have

= 1

Question 21:\lim_{x \to 0} (cosec\hspace{0.1cm}x-cot\hspace{0.1cm}x)

Solution:

In\lim_{x \to 0} (cosec\hspace{0.1cm}x-cot\hspace{0.1cm}x) , as x⇢0

By simplification, we get

\lim_{x \to 0} (\frac{1}{sin\hspace{0.1cm}x}-\frac{cos\hspace{0.1cm}x}{sin\hspace{0.1cm}x})

\lim_{x \to 0} (\frac{1-cos\hspace{0.1cm}x}{sin\hspace{0.1cm}x})

Put x = 0, we get

\lim_{x \to 0} (\frac{1-cos\hspace{0.1cm}x}{sin\hspace{0.1cm}x}) = \frac{0}{0}

As, this limit becomes undefined

Now, let’s simplify the equation to make it equivalent to theorem:

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

By using the trigonometric identities,

cos 2θ = 1-2sin2θ

sin 2θ = 2 sinθ cosθ

Hence, we can write the equation as follows:

\lim_{x \to 0} (\frac{2sin^2(\frac{x}{2})}{2 sin(\frac{x}{2})cos(\frac{x}{2})})

=\lim_{x \to 0} (\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})})

=\lim_{x \to 0} tan(\frac{x}{2})

Putting x=0, we have

= 0

Question 22:\lim_{x \to \frac{\pi}{2}} \frac{tan \hspace{0.1cm}2x}{x-\frac{\pi}{2}}

Solution:

In\lim_{x \to \frac{\pi}{2}} \frac{tan \hspace{0.1cm}2x}{x-\frac{\pi}{2}} , as x⇢\frac{\pi}{2}

Put x =\frac{\pi}{2} , we get

\lim_{x \to \frac{\pi}{2}} \frac{tan \hspace{0.1cm}2x}{x-\frac{\pi}{2}} = \frac{0}{0}

As, this limit becomes undefined

Now, let’s simplify the equation :

Let’s takex-\frac{\pi}{2}=p

As, x⇢\frac{\pi}{2} ⇒ p⇢0

Hence, we can write the equation as follows:

\lim_{p \to 0} \frac{tan \hspace{0.1cm}2(p+\frac{\pi}{2})}{p}

=\lim_{p \to 0} \frac{tan \hspace{0.1cm}(2p+\pi)}{p}

=\lim_{p \to 0} \frac{tan \hspace{0.1cm}(2p)}{p} (As tan (π+θ) = tan θ)

=\lim_{p \to 0} \frac{\frac{sin \hspace{0.1cm}(2p)}{cos\hspace{0.1cm}(2p)}}{p}

=\lim_{p \to 0} \frac{sin \hspace{0.1cm}(2p)}{cos\hspace{0.1cm}(2p) \times p}

Now, let’s multiply and divide the equation by 2 to make it equivalent to theorem

\mathbf{\lim_{x \to 0} \frac{sin x}{x} = 1}

=\lim_{p \to 0} \frac{sin \hspace{0.1cm}(2p)}{cos\hspace{0.1cm}(2p) \times p} \times \frac{2}{2}

=\lim_{p \to 0} \frac{2 sin \hspace{0.1cm}(2p)}{cos\hspace{0.1cm}(2p) \times 2p}

As p⇢0, then 2p⇢0

=2. \lim_{2p \to 0} \frac{sin \hspace{0.1cm}(2p)}{2p} \times \lim_{p \to 0}\frac{1}{cos\hspace{0.1cm}(2p)}

Using the theorem and putting p=0, we have

= 2×1×1

= 2

Question 23: Find\lim_{x \to 0} f(x) and\lim_{x \to 1} f(x) , wheref(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\leq0\\ 3(x+1),\hspace{0.2cm}x>0 \end{cases}

Solution:

Let’s calculate, the limits when x⇢0

Here,

Left limit =\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (2x+3)\\ = 2(0)+3\\ =3

Right limit =\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3(x+1)\\ = 3(0+1)\\ =3

Limit value =\lim_{x \to 0} f(x) = \lim_{x \to 0} (2x+3)\\ = 2(0)+3\\ =3

Hence,\lim_{x \to 0^-} f(x)= \lim_{x \to 0} f(x)=\lim_{x \to 0^+} f(x)=3 , then limit exists

Now, let’s calculate, the limits when x⇢1

Here,

Left limit =\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 3(x+1)\\= 3(1+1)\\=6

Right limit =\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 3(x+1)\\= 3(1+1)\\=6

Limit value =\lim_{x \to 1} f(x) = \lim_{x \to 1} 3(x+1)\\= 3(1+1)\\=6

Hence,\lim_{x \to 1^-} f(x)= \lim_{x \to 1} f(x)=\lim_{x \to 1^+} f(x) = 6 , then limit exists

Question 24: Find\lim_{x \to 1} f(x), wheref(x)= \begin{cases} x^2-1, \hspace{0.2cm}x\leq1\\ -x^2-1,\hspace{0.2cm}x>1 \end{cases}

Solution:

Let’s calculate, the limits when x⇢1

Here,

Left limit =\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (x^2-1)\\= 1^2-1\\=0

Right limit =\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (-x^2-1)\\= -1^2-1\\=-2

As,\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)

Hence, limit does not exists when x⇢1.

Question 25: Evaluate\lim_{x \to 0} f(x) , wheref(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}

Solution:

Let’s calculate, the limits when x⇢0

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{|x|}{x}\\= \frac{-x}{x}\\=-1

Right limit =\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{|x|}{x}\\= \frac{x}{x}\\=1

As,\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)

Hence, limit does not exists when x⇢0.

Question 26: Find \lim_{x \to 0} f(x), wheref(x)= \begin{cases} \frac{x}{|x|}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}

Solution:

Let’s calculate, the limits when x⇢0

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{x}{|x|}\\= \frac{x}{-x}\\=-1

Right limit =\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{x}{|x|}\\= \frac{x}{x}\\=1

As,\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)

Hence, limit does not exists when x⇢0.

Question 27: Find\lim_{x \to 5} f(x) , where f(x)=|x|-5.

Solution:

Let’s calculate, the limits when x⇢5

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Left limit =\lim_{x \to 5^-} f(x) = \lim_{x \to 5^-} |x|-5\\= x-5\\=5-5\\=0

Right limit =\lim_{x \to 5^+} f(x) = \lim_{x \to 5^+} |x|-5\\= x-5\\=5-5\\=0

Hence,\lim_{x \to 5^-} f(x)= \lim_{x \to 5} f(x)=\lim_{x \to 5^+} f(x) = 0 , then limit exists

Question 28: Supposef(x)= \begin{cases} a+bx, \hspace{0.2cm}x<1\\ 4,\hspace{0.2cm}x=1\\ b-ax,\hspace{0.2cm}x>1 \end{cases} and if\lim_{x \to 1} f(x) = f(1) what are possible values of a and b?

Solution:

As, it is given\lim_{x \to 1} f(x) = f(1)

Let’s calculate, the limits when x⇢1

Here,

Left limit =\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} a+bx\\= a+b(1)\\=a+b

Right limit =\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} b-ax\\= b-a(1)\\=b-a

Limit value f(1) = 4

So, as limit exists then it should satisfy

\lim_{x \to 1^-} f(x)= \lim_{x \to 1} f(x)=\lim_{x \to 1^+} f(x) = f(1) = 4

Hence, a+b = 4 and b-a = 4

Solving these equation, we get

a = 0 and b = 4

Question 29: Let a1, a2, . . ., an be fixed real numbers and define a function

f(x) = (x-a1) (x-a2)………… (x-an).

What is\lim_{x \to a_1} f(x) ? For some a ≠ a1, a2, …, an, compute\lim_{x \to a} f(x).

Solution:

Here, f(x) = (x-a1) (x-a2)………… (x-an).

Then,\lim_{x \to a_1} f(x) = \lim_{x \to a_1} (x-a_1) (x-a_2)............ (x-a_n)

=\lim_{x \to a_1} (x-a1) \lim_{x \to a_1}(x-a_2)............ \lim_{x \to a_1}(x-a_n)

= (a1-a1) (a1-a2)………… (a1-an)

\lim_{x \to a_1} f(x) = 0

Now, let’s calculate for\lim_{x \to a} f(x)

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

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

= (a-a1) (a-a2)………… (a-an)

\lim_{x \to a} f(x) = (a-a1) (a-a2)………… (a-an)

Question 30: Iff(x)= \begin{cases} |x|+1, \hspace{0.2cm}x<0\\ 0,\hspace{0.2cm}x=0\\ |x|-1,\hspace{0.2cm}x>0 \end{cases}

For what value (s) of a does\lim_{x \to a} f(x) exists?

Solution:

Here,

As, we know that mod function works differently.

In |x-0|, |x|=x when x>0 and |x|=-x when x<0

Let’s check for three cases of a:

  • When a=0

Let’s calculate, the limits when x⇢0

Left limit =\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (|x|+1)\\= -x+1\\=1

Right limit =\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (|x|-1)\\= x-1\\=-1

As,\lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x)

Hence, limit does not exists when x⇢0.

  • When a>0

Let’s take a=2, for reference

Let’s calculate, the limits when x⇢2

Left limit =\lim_{x \to 2^-} f(x) = \lim_{x \to 2^-} (|x|-1)\\= x-1\\=2-1\\=1

Right limit =\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (|x|-1)\\= x-1\\=2-1\\=1

As,\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x)

Hence, limit exists when x⇢2.

  • When a<0

Let’s take a=-2, for reference

Let’s calculate, the limits when x⇢ -2

Left limit =\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} (|x|+1)\\= x+1\\=-2+1\\=-1

Right limit =\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} (|x|+1)\\= x+1\\=-2+1\\=-1

As,\lim_{x \to -2^-} f(x) = \lim_{x \to -2^+} f(x)

Hence, limit exists when x⇢ -2.

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

Solution:

Here, as it is given

\lim_{x \to 1} \frac{f(x)-2}{x^2-1} = \pi

\frac{\lim_{x \to 1} f(x)-2}{\lim_{x \to 1} x^2-1} = \pi

\lim_{x \to 1} (f(x)-2) = \pi (\lim_{x \to 1} x^2-1)

Put x = 1 in RHS, we get

\lim_{x \to 1} (f(x)-2) = \pi (\lim_{x \to 1} (1^2-1))

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

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

\lim_{x \to 1} f(x) = 2

Hence proved!

Question 32: Iff(x)= \begin{cases} mx^2+n, \hspace{0.2cm}x<0\\ nx+m,\hspace{0.2cm},0\leq x\leq 1\\ nx^3+m, \hspace{0.2cm}x>1 \end{cases} . For what integers m and n does both\lim_{x \to 0} f(x) and\lim_{x \to 1} f(x) exists?

Solution:

Let’s calculate, the limits when x⇢0

Here,

Left limit =\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (mx^2+n)\\ = (m(0)^2+n)\\ =n

Right limit =\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (nx+m)\\ = (n(0)+m)\\ =m

Hence,

\lim_{x \to 0^-} f(x)=\lim_{x \to 0^+} f(x) , then limit exists

m = n

Now, let’s calculate, the limits when x⇢1

Here,

Left limit =\lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} (nx+m)\\= (n(1)+m)\\=n+m

Right limit =\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (nx^3+m)\\= (n(1)^3+m)\\=n+m

Hence,\lim_{x \to 1^-} f(x)=\lim_{x \to 1^+} f(x) = m+n , then limit exists.

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