Anand Classes presents another set of detailed NCERT Solutions for Class 11 Maths Chapter 12 Limits and Derivatives Exercise 12.1 (Set-2) to help students gain deeper understanding and additional practice on limit-related problems. These solutions are prepared according to the latest CBSE and NCERT syllabus, covering various methods of evaluating limits with clear step-by-step explanations. This second set is perfect for students who wish to strengthen their calculus fundamentals and improve problem-solving accuracy. Click the print button to download study material and notes.
NCERT Question 11 : Evaluate:
$$\lim_{x \to 1} \frac{a x^2 + b x + c}{c x^2 + b x + a}, \quad \text{where } a + b + c \ne 0$$
Solution:
As $x \to 1$, substitute $x = 1$ in the expression:
$$\lim_{x \to 1} \frac{a x^2 + b x + c}{c x^2 + b x + a} = \frac{a(1)^2 + b(1) + c}{c(1)^2 + b(1) + a}$$
Simplify:
$$= \frac{a + b + c}{c + b + a} = 1$$
(As it is given that $a + b + c \ne 0$.)
✅ Final Answer:
$$\boxed{1}$$
NCERT Question 12: Evaluate
$$ \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} $$
Solution:
We have
$$ \lim_{x \to -2} \frac{\frac{1}{x} + \frac{1}{2}}{x + 2} $$
First, simplify the numerator:
$$ \frac{1}{x} + \frac{1}{2} = \frac{2 + x}{2x} $$
So the limit becomes
$$ \lim_{x \to -2} \frac{\frac{2 + x}{2x}}{x + 2} = \lim_{x \to -2} \frac{2 + x}{2x(x + 2)} $$
Notice that $(2 + x)$ and $(x + 2)$ are the same term, hence cancel them (for $x \neq -2$):
$$ \lim_{x \to -2} \frac{1}{2x} $$
Now substitute $x = -2$:
$$ \frac{1}{2(-2)} = \frac{1}{-4} = -\frac{1}{4} $$
✅ Final Answer:
$$ \boxed{-\frac{1}{4}} $$
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NCERT Question 13: Evaluate
$$ \lim_{x \to 0} \frac{\sin(ax)}{bx} $$
Solution:
We have
$$ \lim_{x \to 0} \frac{\sin(ax)}{bx} $$
Putting $x = 0$, we get
$$ \frac{\sin(a \times 0)}{b \times 0} = \frac{0}{0} $$
which is an indeterminate form.
To simplify, multiply and divide by $a$ to make it similar to the standard limit theorem
$$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $$
So,
$$ \lim_{x \to 0} \frac{\sin(ax)}{bx} = \lim_{x \to 0} \frac{a}{b} \cdot \frac{\sin(ax)}{ax} $$
Let $t = ax$, then as $x \to 0$, $t \to 0$. Hence,
$$ \lim_{x \to 0} \frac{\sin(ax)}{ax} = \lim_{t \to 0} \frac{\sin t}{t} = 1 $$
Therefore,
$$ \lim_{x \to 0} \frac{\sin(ax)}{bx} = \frac{a}{b} \times 1 $$
✅ Final Answer:
$$ \boxed{\frac{a}{b}} $$
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NCERT Question 14: Evaluate
$$ \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)}, \quad a,b \neq 0 $$
Solution:
We have
$$ \lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)} $$
Putting $x = 0$,
$$ \frac{\sin(a \times 0)}{\sin(b \times 0)} = \frac{0}{0} $$
which is an indeterminate form.
Now, multiply and divide the numerator by $ax$ and the denominator by $bx$ to make it equivalent to the standard limit theorem
$$ \lim_{x \to 0} \frac{\sin x}{x} = 1 $$
Thus,
$$
\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)}
= \lim_{x \to 0}
\frac{\sin(ax)}{ax} \cdot \frac{bx}{\sin(bx)} \cdot \frac{a}{b}
$$
Now separate the limits:
$$
\lim_{x \to 0} \frac{\sin(ax)}{ax} = 1
\quad \text{and} \quad
\lim_{x \to 0} \frac{bx}{\sin(bx)} = 1
$$
Therefore,
$$
\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)}
= \frac{a}{b} \times 1 \times 1
$$
✅ Final Answer:
$$ \boxed{\frac{a}{b}} $$
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NCERT Question 15: Evaluate
$$ \lim_{x \to \pi} \frac{\sin(\pi – x)}{\pi(\pi – x)} $$
Solution:
We have
$$ \lim_{x \to \pi} \frac{\sin(\pi – x)}{\pi(\pi – x)} $$
Substituting $x = \pi$,
$$
\frac{\sin(\pi – \pi)}{\pi(\pi – \pi)} = \frac{0}{0}
$$
which is an indeterminate form.
Let
$$ p = \pi – x $$
Then, as $x \to \pi$, we have $p \to 0$.
Thus,
$$
\lim_{x \to \pi} \frac{\sin(\pi – x)}{\pi(\pi – x)}
= \lim_{p \to 0} \frac{\sin p}{\pi p}
$$
Now, recall the standard limit theorem:
$$ \lim_{p \to 0} \frac{\sin p}{p} = 1 $$
Applying this,
$$
\lim_{p \to 0} \frac{\sin p}{\pi p}
= \frac{1}{\pi} \lim_{p \to 0} \frac{\sin p}{p}
= \frac{1}{\pi} \times 1
$$
✅ Final Answer:
$$ \boxed{\frac{1}{\pi}} $$
Master calculus limits and NCERT Class 11 Maths exercises with Anand Classes — ideal for JEE Main, CUET, and NDA aspirants.
NCERT Question 16 : Evaluate
$$\lim_{x \to 0} \frac{\cos x}{\pi – x}$$
Solution :
In
$$\lim_{x \to 0} \frac{\cos x}{\pi – x},$$
as $x \to 0$
Substitute $x = 0$, we get
$$\frac{\cos(0)}{\pi – 0} = \frac{1}{\pi}$$
Hence,
$$\boxed{\frac{1}{\pi}}$$
NCERT Question 17 : Evaluate
$$\lim_{x \to 0} \frac{\cos 2x – 1}{\cos x – 1}$$
Solution :
In
$$\lim_{x\to 0}\frac{\cos 2x – 1}{\cos x – 1},\quad \text{as }x\to 0$$
As we know, $\cos 2\theta = 1 – 2\sin^2\theta$.
Substituting the values, we get
$$
\lim_{x\to 0}\frac{1-2\sin^2 x – 1}{1-2\sin^2\left(\tfrac{x}{2}\right)-1}
$$
which simplifies to
$$
\lim_{x\to 0}\frac{-2\sin^2 x}{-2\sin^2\left(\tfrac{x}{2}\right)}
= \lim_{x\to 0}\frac{\sin^2 x}{\sin^2\left(\tfrac{x}{2}\right)}.
$$
Put $x=0$, the expression is of the indeterminate form $\tfrac{0}{0}$, so we use the standard limit
$$\lim_{u\to 0}\frac{\sin u}{u}=1.$$
Rewrite the ratio using factors that match that theorem:
$$
\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^2 \times \left(\frac{x}{\tfrac{x}{2}}\right)^2
= \left(\lim_{x\to 0}\frac{\sin x}{x}\right)^2 \times \left(\lim_{x\to 0}\frac{x}{\tfrac{x}{2}}\right)^2.
$$
Using $\lim_{x\to 0}\dfrac{\sin x}{x}=1$ and simplifying the second factor:
$$
= (1)^2 \times \left(\frac{x}{x/2}\right)^2
= 1 \times (2)^2
= 4.
$$
Final Answer
$$
\boxed{4}
$$
NCERT Question 19 : Evaluate
$$\lim_{x\to 0} x\sec x$$
Solution:
In
$$\lim_{x\to 0} x\sec x,\quad \text{as }x\to 0$$
Put $x=0$, we get
$$\lim_{x\to 0} x\sec x = 0\times 1$$
$$=0$$
NCERT Question 20
$$\lim_{x\to 0}\frac{bx+\sin ax}{\sin bx+ax}$$
Solution:
In
$$\lim_{x\to 0}\frac{bx+\sin ax}{\sin bx+ax},\quad \text{as }x\to 0$$
Put $x=0$, we get
$$\lim_{x\to 0}\frac{bx+\sin ax}{\sin bx+ax}=\frac{0}{0}$$
As this limit becomes undefined, now let’s simplify the expression to make it equivalent to the standard theorem
$$\lim_{u\to 0}\frac{\sin u}{u}=1.$$
Hence, we can write the equation as follows:
$$
\lim_{x\to 0}
\frac{\dfrac{\sin(ax)}{ax}\times ax + bx}{ax + \dfrac{\sin(bx)}{bx}\times bx}
$$
Breaking into factors that use the standard limit:
$$\lim_{x\to 0}\frac{\dfrac{\sin(ax)}{ax}\times ax + bx}{ax + \dfrac{\sin(bx)}{bx}\times bx}=
\frac{\displaystyle\lim_{x\to 0}\frac{\sin(ax)}{ax}\times\lim_{x\to 0}ax + \lim_{x\to 0}bx}
{\displaystyle\lim_{x\to 0}ax + \lim_{x\to 0}\frac{\sin(bx)}{bx}\times\lim_{x\to 0}bx}$$
By using the theorem, we get
$$
= \frac{1\times\lim_{x\to 0}ax + \lim_{x\to 0}bx}{\lim_{x\to 0}ax + 1\times\lim_{x\to 0}bx}
$$
$$
= \frac{\lim_{x\to 0}(ax+bx)}{\lim_{x\to 0}(ax+bx)}
= \lim_{x\to 0}1 = 1
$$
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