Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 2

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 take $x-\frac{\pi}{2}=p$
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 $\lim_{x \to 1} f(x)$ , where $f(x)= \begin{cases} 2x+3, \hspace{0.2cm}x\leq0\\ 3(x+1),\hspace{0.2cm}x>0 \end{cases}$

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 $\lim_{x \to 1} f(x)$ , where $f(x)= \begin{cases} x^2-1, \hspace{0.2cm}x\leq1\\ -x^2-1,\hspace{0.2cm}x>1 \end{cases}$

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 $\lim_{x \to 0} f(x)$ , where $f(x)= \begin{cases} \frac{|x|}{x}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}$

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 $\lim_{x \to 0} f(x)$ , where $f(x)= \begin{cases} \frac{x}{|x|}, \hspace{0.2cm}x\neq0\\ 0,\hspace{0.2cm}x=0 \end{cases}$

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: Suppose $f(x)= \begin{cases} a+bx, \hspace{0.2cm}x<1\\ 4,\hspace{0.2cm}x=1\\ b-ax,\hspace{0.2cm}x>1 \end{cases}$ 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, . . ., a_n 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-a_n).
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-a_n)
$\lim_{x \to a_1} f(x)$ = 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-a_n)
$\lim_{x \to a} f(x)$ = (a-a₁) (a-a₂)………… (a-a_n)

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

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: If $f(x)= \begin{cases} mx^2+n, \hspace{0.2cm}x<0\\ nx+m,\hspace{0.2cm},0\leq x\leq 1\\ nx^3+m, \hspace{0.2cm}x>1 \end{cases}$ . 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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Last Updated : 07 Apr, 2021

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Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 2

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

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

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

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

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

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

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

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

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

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

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

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

Question 29: Let a₁, a₂, . . ., a_n 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 $\lim_{x \to a} f(x)$ .

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

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

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

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

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Class 11 NCERT Solutions- Chapter 13 Limits And Derivatives – Exercise 13.1 | Set 2

Question 17:

Question 18:

Question 19:

Question 20:

Question 21:

Question 22:

Question 23: Findand, where

Question 24: Find, where

Question 25: Evaluate, where

Question 26: Find , where

Question 27: Find, where f(x)=|x|-5.

Question 28: Supposeand ifwhat are possible values of a and b?

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

f(x) = (x-a1) (x-a2)………… (x-an).

What is? For some a ≠ a1, a2, …, an, compute.

Question 30: If

For what value (s) of a doesexists?

Question 31: If the function f(x) satisfies, evaluate

Question 32: If. For what integers m and n does bothandexists?

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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

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

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

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

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

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

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

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

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

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

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

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

Question 29: Let a₁, a₂, . . ., a_n 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 $\lim_{x \to a} f(x)$ .

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

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

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

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