Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 1

Evaluate the following limits:

Question 1. $\lim_{x\to0}\frac{5^x-1}{\sqrt{4+x}-2}$

Solution:

We have,
= $\lim_{x\to0}\frac{5^x-1}{\sqrt{4+x}-2}$
= $\lim_{x\to0}\frac{(5^x-1)(\sqrt{4+x}+2)}{(\sqrt{4+x}-2)(\sqrt{4+x}+2)}$
= $\lim_{x\to0}\frac{(5^x-1)(\sqrt{4+x}+2)}{4+x-4}$
= $\lim_{x\to0}\frac{(5^x-1)(\sqrt{4+x}+2)}{x}$
= $\lim_{x\to0}\frac{(5^x-1)}{x}\lim_{x\to0}(\sqrt{4+x}+2)$
= $4\lim_{x\to0}\frac{(5^x-1)}{x}$
We know, $lim_{x\to0}\frac{a^x-1}{x}=loga$ . So, we have,
= 4 log 5

Question 2. $\lim_{x\to0}\frac{log(1+x)}{3^x-1}$

Solution:

We have,
= $\lim_{x\to0}\frac{log(1+x)}{3^x-1}$
= $\lim_{x\to0}\frac{log(1+x)}{x}×\frac{1}{\lim_{x\to0}\frac{3^x-1}{x}}$
We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ . So, we have,
= $\frac{1}{\lim_{x\to0}\frac{3^x-1}{x}}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= $\frac{1}{log3}$

Question 3. $\lim_{x\to0}\frac{a^x+a^{-x}-2}{x^2}$

Solution:

We have,
= $\lim_{x\to0}\frac{a^x+a^{-x}-2}{x^2}$
= $\lim_{x\to0}\frac{a^{2x}+1-2a^x}{x^2}$
= $\lim_{x\to0}(\frac{a^x-1}{x})^2×\frac{1}{a^x}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= $(log_ea)^2×\frac{1}{a^0}$
= (log_e a)²

Question 4. $\lim_{x\to0}\frac{a^{mx}-1}{b^{nx}-1}$ , n≠0

Solution:

We have,
= $\lim_{x\to0}\frac{a^{mx}-1}{b^{nx}-1}$
= $\lim_{x\to0}\frac{a^{mx}-1}{mx}×\frac{1}{\lim_{x\to0}\frac{b^{nx}-1}{nx}}×\frac{m}{n}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= $loga×\frac{1}{logb}×\frac{m}{n}$
= $\frac{mloga}{nlogb}$

Question 5. $\lim_{x\to0}\frac{a^x+b^x-2}{x}$

Solution:

We have,
= $\lim_{x\to0}\frac{a^x+b^x-2}{x}$
= $\lim_{x\to0}\frac{(a^x-1)+(b^x-1)}{x}$
= $\lim_{x\to0}\left(\frac{a^x-1}{x}+\frac{b^x-1}{x}\right)$
= $\lim_{x\to0}\frac{a^x-1}{x}+\lim_{x\to0}\frac{b^x-1}{x}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= log a + log b
= log (ab)

Question 6. $lim_{x\to0}(\frac{9^x-2.6^x+4^x}{x^2})$

Solution:

We have,
= $lim_{x\to0}(\frac{9^x-2.6^x+4^x}{x^2})$
= $lim_{x\to0}\frac{(3^x)^2-2.6^x+(2^x)^2}{x^2}$
= $lim_{x\to0}\left(\frac{3^x-2^x}{x}\right)^2$
= $lim_{x\to0}\left(\frac{3^x-1-(2^x-1)}{x}\right)^2$
= $lim_{x\to0}\left(\frac{3^x-1}{x}-\frac{2^x-1}{x}\right)^2$
= $\left(lim_{x\to0}\frac{3^x-1}{x}-lim_{x\to0}\frac{2^x-1}{x}\right)^2$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= (log 3 − log 2)²
= $\left(log\frac{3}{2}\right)^2$

Question 7. $\lim_{x\to0}\frac{8^x-4^x-2^x+1}{x^2}$

Solution:

We have,
= $\lim_{x\to0}\frac{8^x-4^x-2^x+1}{x^2}$
= $\lim_{x\to0}\frac{(2^x-1)^2(2^x+1)}{x^2}$
= $\lim_{x\to0}\frac{(2^x-1)^2}{{x^2}}×\lim_{x\to0}(2^x+1)$
= $2\lim_{x\to0}\left(\frac{2^x-1}{x}\right)^2$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= 2(log2)²

Question 8. $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{x}$

Solution:

We have,
= $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{x}$
= $\lim_{x\to0}\frac{(a^{mx}-1)-(b^{nx}-1)}{x}$
= $\lim_{x\to0}\left(\frac{a^{mx}-1}{x}-\frac{b^{nx}-1}{x}\right)$
= $\lim_{x\to0}\left(m×\frac{a^{mx}-1}{mx}\right)-\lim_{x\to0}\left(n×\frac{b^{nx}-1}{nx}\right)$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= m log a − n log b
= log a^m − log bⁿ
= $log\left(\frac{a^m}{b^n}\right)$

Question 9. $\lim_{x\to0}\frac{a^x+b^x+c^x-3}{x}$

Solution:

We have,
= $\lim_{x\to0}\frac{a^x+b^x+c^x-3}{x}$
= $\lim_{x\to0}\frac{(a^x-1)+(b^x-1)+(c^x-1)}{x}$
= $\lim_{x\to0}\frac{a^x-1}{x}+\lim_{x\to0}\frac{b^x-1}{x}+\lim_{x\to0}\frac{c^x-1}{x}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= log a + log b + log c
= log (abc)

Question 10. $\lim_{x\to2}\frac{x-2}{log_a(x-1)}$

Solution:

We have,
= $\lim_{x\to2}\frac{x-2}{log_a(x-1)}$
Let x − 2 = h. So, we get,
= $\lim_{h\to0}\frac{h}{log_a(h+1)}$
= $\lim_{h\to0}\frac{h}{\frac{log(h+1)}{loga}}$
= $\lim_{h\to0}\frac{loga}{\frac{log(h+1)}{h}}$
= $\frac{loga}{\lim_{h\to0}\frac{log(h+1)}{h}}$
We know, $lim_{x\to{a}}\frac{log(1+x)}{x}=1$ . So, we have,
= log a

Question 11. $\lim_{x\to0}\frac{5^x+3^x+2^x-3}{x}$

Solution:

We have,
= $\lim_{x\to0}\frac{5^x+3^x+2^x-3}{x}$
= $\lim_{x\to0}\frac{(5^x-1)+(3^x-1)+(2^x-1)}{x}$
= $\lim_{x\to0}\frac{5^x-1}{x}+\lim_{x\to0}\frac{4^x-1}{x}+\lim_{x\to0}\frac{3^x-1}{x}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= log 5 + log 3 + log 2
= log (5×3×2)
= log 30

Question 12. $\lim_{x\to0}(a^{\frac{1}{x}}-1)x$

Solution:

We have,
= $\lim_{x\to0}(a^{\frac{1}{x}}-1)x$
Let 1/x = h. We get,
= $\lim_{h\to0}(a^h-1)\frac{1}{h}$
= $\lim_{h\to0}\frac{a^h-1}{h}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= log a

Question 13. $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{sinkx}$

Solution:

We have,
= $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{sinkx}$
= $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{kx×\frac{sinkx}{kx}}$
= $\frac{1}{k}\lim_{x\to0}\frac{\frac{a^{mx}-b^{nx}}{x}}{\frac{sinkx}{kx}}$
= $\frac{1}{k}\lim_{x\to0}\frac{\frac{a^{mx}-1-(b^{nx}-1)}{x}}{\frac{sinkx}{kx}}$
= $\frac{1}{k}\lim_{x\to0}\frac{\frac{a^{mx}-1}{x}-\frac{b^{nx}-1}{x}}{\frac{sinkx}{kx}}$
= $\frac{1}{k}\lim_{x\to0}\frac{m×\frac{a^{mx}-1}{mx}-n×\frac{b^{nx}-1}{nx}}{\frac{sinkx}{kx}}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ and $lim_{x\to0}\frac{sinx}{x}=1$ , we get,
= $\frac{mloga-nlogb}{k}$
= $\frac{loga^m-logb^n}{k}$
= $\frac{1}{k}log(\frac{a^m}{b^n})$

Question 14. $\lim_{x\to0}\frac{a^x+b^x-c^x-d^x}{x}$

Solution:

We have,
= $\lim_{x\to0}\frac{a^x+b^x-c^x-d^x}{x}$
= $\lim_{x\to0}\frac{(a^x-1)+(b^x-1)-(c^x-1)-(d^x-1)}{x}$
= $\lim_{x\to0}\frac{a^x-1}{x}+\lim_{x\to0}\frac{b^x-1}{x}-\lim_{x\to0}\frac{c^x-1}{x}-\lim_{x\to0}\frac{d^x-1}{x}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,
= log a + log b − log c − log d
= log a+ log b − (log c + log d)
= log ab − log cd
= $\frac{ab}{cd}$

Question 15. $\lim_{x\to0}\frac{e^x-1+sinx}{x}$

Solution:

We have,
= $\lim_{x\to0}\frac{e^x-1+sinx}{x}$
= $\lim_{x\to0}\frac{e^x-1}{x}+\lim_{x\to0}\frac{sinx}{x}$
As $lim_{x\to0}\frac{a^x-1}{x}=loga$ and $lim_{x\to0}\frac{sinx}{x}=1$ , we get,
= log e + 1
= 1 + 1
= 2

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Last Updated : 04 May, 2021

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Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 1

Evaluate the following limits:

Question 1. $\lim_{x\to0}\frac{5^x-1}{\sqrt{4+x}-2}$

Question 2. $\lim_{x\to0}\frac{log(1+x)}{3^x-1}$

Question 3. $\lim_{x\to0}\frac{a^x+a^{-x}-2}{x^2}$

Question 4. $\lim_{x\to0}\frac{a^{mx}-1}{b^{nx}-1}$ , n≠0

Question 5. $\lim_{x\to0}\frac{a^x+b^x-2}{x}$

Question 6. $lim_{x\to0}(\frac{9^x-2.6^x+4^x}{x^2})$

Question 7. $\lim_{x\to0}\frac{8^x-4^x-2^x+1}{x^2}$

Question 8. $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{x}$

Question 9. $\lim_{x\to0}\frac{a^x+b^x+c^x-3}{x}$

Question 10. $\lim_{x\to2}\frac{x-2}{log_a(x-1)}$

Question 11. $\lim_{x\to0}\frac{5^x+3^x+2^x-3}{x}$

Question 12. $\lim_{x\to0}(a^{\frac{1}{x}}-1)x$

Question 13. $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{sinkx}$

Question 14. $\lim_{x\to0}\frac{a^x+b^x-c^x-d^x}{x}$

Question 15. $\lim_{x\to0}\frac{e^x-1+sinx}{x}$

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

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Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 1

Evaluate the following limits:

Question 1.

Question 2.

Question 3.

Question 4. , n≠0

Question 5.

Question 6.

Question 7.

Question 8.

Question 9.

Question 10.

Question 11.

Question 12.

Question 13.

Question 14.

Question 15.

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 1. $\lim_{x\to0}\frac{5^x-1}{\sqrt{4+x}-2}$

Question 2. $\lim_{x\to0}\frac{log(1+x)}{3^x-1}$

Question 3. $\lim_{x\to0}\frac{a^x+a^{-x}-2}{x^2}$

Question 4. $\lim_{x\to0}\frac{a^{mx}-1}{b^{nx}-1}$ , n≠0

Question 5. $\lim_{x\to0}\frac{a^x+b^x-2}{x}$

Question 6. $lim_{x\to0}(\frac{9^x-2.6^x+4^x}{x^2})$

Question 7. $\lim_{x\to0}\frac{8^x-4^x-2^x+1}{x^2}$

Question 8. $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{x}$

Question 9. $\lim_{x\to0}\frac{a^x+b^x+c^x-3}{x}$

Question 10. $\lim_{x\to2}\frac{x-2}{log_a(x-1)}$

Question 11. $\lim_{x\to0}\frac{5^x+3^x+2^x-3}{x}$

Question 12. $\lim_{x\to0}(a^{\frac{1}{x}}-1)x$

Question 13. $\lim_{x\to0}\frac{a^{mx}-b^{nx}}{sinkx}$

Question 14. $\lim_{x\to0}\frac{a^x+b^x-c^x-d^x}{x}$

Question 15. $\lim_{x\to0}\frac{e^x-1+sinx}{x}$