GSEB Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.1

Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.1 Textbook Questions and Answers.

Gujarat Board Textbook Solutions Class 11 Maths Chapter 13 Limits and Derivatives Ex 13.1

Evalute the following limits in questions 1 to 22:
1. \(\lim _{x \rightarrow 3}\) (x + 3)
2. \(\lim _{x \rightarrow π}\) (x – \(\frac{22}{7}\))
3. \(\lim _{r \rightarrow 1}\) (πr²)
4. \(\lim _{x \rightarrow 4}\) \(\frac{4x+3}{x-2}\)
5. \(\lim _{x \rightarrow -1}\) \(\frac{x^{10}+x^{5}+1}{x-1}\)
6. \(\lim _{x \rightarrow 0}\) \(\frac{(x+1)^{5}-1}{x}\)
7. \(\lim _{x \rightarrow 2}\) \(\frac{3 x^{2}-x-10}{x^{2}-4}\)
8. \(\lim _{x \rightarrow 3}\) \(\frac{x^{4}-81}{2x^{2}-5x-3}\)
9. \(\lim _{x \rightarrow 0}\) \(\frac{ax+b}{cx+1}\)
10. \(\lim _{z \rightarrow 1}\) \(\frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1}\)
11. \(\lim _{x \rightarrow 1}\) \(\frac{ax^{2}+bx+c}{cx^{2}+bx+a}\)
12. \(\lim _{x \rightarrow -2}\) \(\frac{1/x+1/2}{x+2}\)
13. \(\lim _{x \rightarrow 0}\) \(\frac{sin ax}{bx}\)
14. \(\lim _{x \rightarrow 0}\) \(\frac{sin ax}{bx}\), ab ≠ 0
15. \(\lim _{x \rightarrow π}\) \(\frac{sin (π-x)}{π(π-x)}\)
16. \(\lim _{x \rightarrow 0}\) \(\frac{cos x}{π-x}\)
17. \(\lim _{x \rightarrow 0}\) \(\frac{cos 2x-1}{cos x-1}\)
18. \(\lim _{x \rightarrow 0}\) \(\frac{ax+xcosx}{bsinx}\)
19. \(\lim _{x \rightarrow 0}\) (x sec x)
20. \(\lim _{x \rightarrow 0}\) \(\frac{sin ax+bx}{ax+sinbx}\), a, b, a + b ≠ 0
21. \(\lim _{x \rightarrow 0}\) (cosec x – cot x)
22. \(\lim _{x \rightarrow \frac{\pi}{2}} \frac{\tan 2 \pi}{x-\frac{\pi}{2}}\)
Solutions to questions 1 to 22:
1. \(\lim _{x \rightarrow 3}\) (x + 3) = 3 + 3 = 6

2. \(\lim _{x \rightarrow π}\) (x – \(\frac{22}{7}\)) = π – \(\frac{22}{7}\)

3. \(\lim _{r \rightarrow 1}\) (πr²) = π.1² = π.

4. \(\lim _{x \rightarrow 4}\) \(\frac{4x+3}{x-2}\) = \(\frac{4×4+3}{4-2}\) = \(\frac{16+3}{2}\) = \(\frac{19}{2}\).

5.

6.

7.

8.

9. \(\lim _{x \rightarrow 0}\) \(\frac{ax+b}{cx+1}\) = \(\frac{0+b}{0+1}\) = \(\frac{b}{1}\) = b.

10.

= (1)^1/6 + 1
= 1 + 1 = 2.

11.

12.

13. \(\lim _{x \rightarrow 0}\) \(\frac{sin ax}{bx}\) = \(\lim _{x \rightarrow 0}\). \(\frac{sin ax}{b.ax}\).a = \(\lim _{x \rightarrow 0}\) \(\frac{sin ax}{ax}\). \(\frac{a}{b}\) ) = \(\frac{a}{b}\).

14.

15. \(\lim _{x \rightarrow π}\) \(\frac{sin (π-x)}{π(π-x)}\).
Put π – x = θ, As x → π, θ → 0 (zero)

16. \(\lim _{x \rightarrow 0}\) \(\frac{cos x}{π-x}\) = \(\frac{cos 0}{π-0}\) = \(\frac{1}{π}\).

17.

18.

19.

20.

21.

22.

Question 23.
Find \(\lim _{x \rightarrow 0}\) f(x) and \(\lim _{x \rightarrow 1}\) f(x), where
f(x) = \(\left\{\begin{array}{l}
2 x+3, x \leq 0 \\
3(x+1), x>0
\end{array}\right.\)
Solution:
To find \(\lim _{x \rightarrow 0^{-}}\) f(x), we have to find \(\lim _{x \rightarrow 0^{-}}\) (2x + 3).
Here, we put values of x near the zero but less than zero, i.e;

Alternative method:

Question 24.
Find \(\lim _{x \rightarrow 1}\) f(x), where f(x) = \(\left\{\begin{array}{l}
x^{2}-1, x \leq 1 \\
-x^{2}-1, x>1
\end{array}\right.\)
Solution:
When x ≤ 1, f(x) = x² – 1:

∴ \(\lim _{x \rightarrow 1}\) f(x) does not exist.

Question 25.
Evalute \(\lim _{x \rightarrow 0}\) f(x), where f(x) = \(\left\{\begin{array}{l}
\frac{|x|}{x}, x \neq 0 \\
0, \quad x=0
\end{array}\right.\).
Solution:
When x < 0, |x| = – x.

i.e; \(\lim _{x \rightarrow 0}\) f(x) does not exist.

Question 26.
Find \(\lim _{x \rightarrow 0}\) f(x), where f(x) = \(\left\{\begin{array}{l}
\frac{x}{|x|}, x \neq 0 \\
0, x=0
\end{array}\right.\).
Solution:
When x < 0, |x| = – x image 19 ∴ \(\lim _{x \rightarrow 0}\) f(x) does not exist.

Question 27.
Find \(\lim _{x \rightarrow 5}\) f(x), where f(x) = |x| – 5.
Solution: When x > 5, put x = 5 + h, where h is small
So, |x| = |5 + h| = 5 + h.

Question 28.
Suppose f(x) = \(\left\{\begin{array}{l}
a+b x, x<1 \\ 4, x=1 \\ b-a x, x>1
\end{array}\right.\) and if \(\lim _{x \rightarrow 1}\) f(x) = f(1), what are possible values of a and b?
Solution:
When x < 1, f(x) = a + bx

Adding (1) and (2), we get
2b = 8 ⇒ b = 4
and a = 0.
Then, a = 0 and b = 4.

Question 29.
Let a₁, a₂, ………………. a_n be fixed real numbers and define a
f(x) = (x – a₁)(x – a₂) ………………… (x – a_n).
What is the \(\lim _{x \rightarrow a_{1}}\) f(x)? For some a ≠ a₁, a₂, ………….., a_n, compute \(\lim _{x \rightarrow a}\) f(x).
Solution:
(i) Consider the factor x – a₁, As x → a₁, x – a₁ → 0

When a ≠ a₁, a₂, a₃, …………………
Consider the factor x – a₁, As x → a, x – a₁ → a – a₁.
a – a₁ is neither zero nor indeterminate. It is a unique value. So, is the case with other factors and their values, the (a – a₂)₁ (a – a₃), ………….. (a – a_n)
∴ \(\lim _{x \rightarrow a}\) f(x) = \(\lim _{x \rightarrow a}\) (x – a₁)(x – a₂) …………. : (x – a_n)
= (a – a₁)(a – a₂) …………… (a – a_n).

Question 30.
If f(x) = \(\left\{\begin{array}{l}
|x|+1, x<0 \\ 0, x=0 \\ |x|-1, x>0
\end{array}\right.\), then
Solution:
(i) consider the limit at x = 0

∴ \(\lim _{x \rightarrow 0}\) f(x) does not exist.

(ii) When x ≠ 0, let x = a (a < 0)

At all points x < a, \(\lim _{x \rightarrow a}\) f(x) exists. (iii) When a > 0, f(x) = |x| – 1 = x – 1

Thus, \(\lim _{x \rightarrow a}\) f(x) does not exist and it is equal to a – 1.
When a < 0, \(\lim _{x \rightarrow a}\) f(x) = 1 – a. When a > 0, \(\lim _{x \rightarrow a}\) f(x) = a – 1.

Question 31.
If the function f(x) satisfies \(\lim _{x \rightarrow 1}\) \(\frac{f(x)-2}{x^{2}-1}\) = π, evaluate \(\lim _{x \rightarrow 1}\) f(x).
Solution:

Question 32.
For what integers m and n does both \(\lim _{x \rightarrow 0}\) f(x) and \(\lim _{x \rightarrow 1}\) f(x) exist for the following function?

Solution:
(i) When x < 0, f(x) = mx² + n