Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 2

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Evaluate the following limits:

Question 16. $\lim_{x\to0}\frac{sin2x}{e^x-1}$

Solution:

We have,

= $\lim_{x\to0}\frac{sin2x}{e^x-1}$

= $\lim_{x\to0}\frac{2sin2x}{2x}×\lim_{x\to0}\frac{x}{e^x-1}$

= $\lim_{x\to0}\frac{2sin2x}{2x}×\lim_{x\to0}\frac{1}{\frac{e^x-1}{x}}$

As $lim_{x\to0}\frac{a^x-1}{x}=loga$ and $lim_{x\to0}\frac{sinx}{x}=1$ , we get,

= $2×\frac{1}{loge}$

= 2

Question 17. $\lim_{x\to0}\frac{e^{sinx}-1}{x}$

Solution:

We have,

= $\lim_{x\to0}\frac{e^{sinx}-1}{x}$

= $\lim_{x\to0}\left(\frac{e^{sinx}-1}{sinx}×\frac{sinx}{x}\right)$

= $\lim_{x\to0}\left(\frac{e^{sinx}-1}{sinx}\right)×\lim_{x\to0}\left(\frac{sinx}{x}\right)$

As $lim_{x\to0}\frac{a^x-1}{x}=loga$ and $lim_{x\to0}\frac{sinx}{x}=1$ , we get,

= log e × 1

= 1

Question 18. $\lim_{x\to0}\frac{e^{2x}-e^x}{sin2x}$

Solution:

We have,

= $\lim_{x\to0}\frac{e^{2x}-e^x}{sin2x}$

= $\lim_{x\to0}\frac{e^{2x}-1-(e^x-1)}{sin2x}$

= $\lim_{x\to0}\left(\frac{e^{2x}-1}{sin2x}-\frac{e^x-1}{sin2x}\right)$

= $\lim_{x\to0}\left(\frac{e^{2x}-1}{sin2x}\right)-\lim_{x\to0}\left(\frac{e^x-1}{sin2x}\right)$

= $\lim_{x\to0}\left(\frac{\frac{e^{2x}-1}{2x}}{\frac{sin2x}{2x}}\right)-\lim_{x\to0}\left(\frac{\frac{e^x-1}{2x}}{\frac{sin2x}{2x}}\right)$

As $lim_{x\to0}\frac{a^x-1}{x}=loga$ and $lim_{x\to0}\frac{sinx}{x}=1$ , we get,

= 1 − $\frac{1}{2}$

= $\frac{1}{2}$

Question 19. $\lim_{x\to0}\frac{logx-loga}{x-a}$

Solution:

We have,

= $\lim_{x\to0}\frac{logx-loga}{x-a}$

= $\lim_{x\to0}\frac{log\frac{x}{a}}{x-a}$

= $\lim_{x\to0}\frac{log\frac{x}{a}}{a(\frac{x}{a}-1)}$

Let h = x/a − 1. We get,

= $\lim_{x\to0}\frac{log(h+1)}{ah}$

We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ . So, we have,

= $\frac{1}{a}\lim_{x\to0}\frac{log(h+1)}{h}$

= $\frac{1}{a}$

Question 20. $\lim_{x\to0}\frac{log(a+x)-log(a-x)}{x}$

Solution:

We have,

= $\lim_{x\to0}\frac{log(a+x)-log(a-x)}{x}$

= $\lim_{x\to0}\frac{log(\frac{a+x}{a-x})}{x}$

= $\lim_{x\to0}\frac{log(\frac{a-x+2x}{a-x})}{x}$

= $\lim_{x\to0}\frac{log(1+\frac{2x}{a-x})}{x}$

= $\lim_{x\to0}\left(\frac{log(1+\frac{2x}{a-x})}{\frac{2x}{a-x}}×\frac{2}{a-x}\right)$

= $\lim_{x\to0}\left(\frac{log(1+\frac{2x}{a-x})}{\frac{2x}{a-x}}\right)×\lim_{x\to0}\left(\frac{2}{a-x}\right)$

We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ . So, we have,

= $\lim_{x\to0}\left(\frac{2}{a-x}\right)$

= $\frac{2}{a}$

Question 21. $\lim_{x\to0}\frac{log(2+x)+log(0.5)}{x}$

Solution:

We have,

= $\lim_{x\to0}\frac{log(2+x)+log(0.5)}{x}$

= $\lim_{x\to0}\frac{log(2+x)+log(\frac{1}{2})}{x}$

= $\lim_{x\to0}\frac{log(1+\frac{x}{2})}{x}$

= $\lim_{x\to0}\frac{log(1+\frac{x}{2})}{2×\frac{x}{2}}$

We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ . So, we have,

= $\frac{1}{2}$

Question 22. $\lim_{x\to0}\frac{log(a+x)-loga}{x}$

Solution:

We have,

= $\lim_{x\to0}\frac{log(a+x)-loga}{x}$

= $\lim_{x\to0}\frac{log(\frac{a+x}{a})}{x}$

= $\lim_{x\to0}\frac{log(1+\frac{x}{a})}{x}$

= $\lim_{x\to0}\frac{log(1+\frac{x}{a})}{\frac{x}{a}×a}$

We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ . So, we have,

= $\frac{1}{a}$

Question 23. $\lim_{x\to0}\frac{log(3+x)-log(3-x)}{x}$

Solution:

We have,

= $\lim_{x\to0}\frac{log(3+x)-log(3-x)}{x}$

= $\lim_{x\to0}\frac{log(\frac{3+x}{3-x})}{x}$

= $\lim_{x\to0}\frac{log(\frac{3-x+2x}{3-x})}{x}$

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

= $\lim_{x\to0}\left(\frac{log(1+\frac{2x}{3-x})}{\frac{2x}{3-x}}×\frac{2}{3-x}\right)$

= $\lim_{x\to0}\left(\frac{log(1+\frac{2x}{3-x})}{\frac{2x}{3-x}}\right)×\lim_{x\to0}\left(\frac{2}{3-x}\right)$

We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ . So, we have,

= $\lim_{x\to0}\left(\frac{2}{3-x}\right)$

= $\frac{2}{3}$

Question 24. $\lim_{x\to0}\frac{8^x-2^x}{x}$

Solution:

We have,

= $\lim_{x\to0}\frac{8^x-2^x}{x}$

= $\lim_{x\to0}\frac{8^x-1-(2^x-1)}{x}$

= $\lim_{x\to0}\left(\frac{8^x-1}{x}-\frac{2^x-1}{x}\right)$

= $\lim_{x\to0}\left(\frac{8^x-1}{x}\right)-\lim_{x\to0}\left(\frac{2^x-1}{x}\right)$

As $lim_{x\to0}\frac{a^x-1}{x}=loga$ , we get,

= log 8 − log 2

= $log(\frac{8}{2})$

= log 4

Question 25. $\lim_{x\to0}\frac{x(2^x-1)}{cosx}$

Solution:

We have,

= $\lim_{x\to0}\frac{x(2^x-1)}{cosx}$

= $\lim_{x\to0}\frac{x(2^x-1)}{2sin^2\frac{x}{2}}$

= $\lim_{x\to0}\left(\frac{2^x-1}{x}\right)×\lim_{x\to0}\left(\frac{x^2}{\frac{x^2}{2}\left(\frac{sin\frac{x}{2}}{\frac{x}{2}}\right)^2}\right)$

As $lim_{x\to0}\frac{a^x-1}{x}=loga$ and $lim_{x\to0}\frac{sinx}{x}=1$ , we get,

= (log 2) × 2

= 2 log 2

= log 4

Question 26. $\lim_{x\to0}\frac{\sqrt{1+x}-1}{log(1+x)}$

Solution:

We have,

= $\lim_{x\to0}\frac{\sqrt{1+x}-1}{log(1+x)}$

= $\lim_{x\to0}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{log(1+x)(\sqrt{1+x}+1)}$

= $\lim_{x\to0}\frac{1+x-1}{log(1+x)(\sqrt{1+x}+1)}$

= $\lim_{x\to0}\frac{x}{log(1+x)(\sqrt{1+x}+1)}$

= $\lim_{x\to0}\frac{1}{\frac{log(1+x)}{x}(\sqrt{1+x}+1)}$

= $\lim_{x\to0}\frac{1}{\frac{log(1+x)}{x}}×\lim_{x\to0}\frac{1}{(\sqrt{1+x}+1)}$

We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ . So, we have,

= $\frac{1}{1+1}$

= $\frac{1}{2}$

Question 27. $\lim_{x\to0}\frac{log|1+x^3|}{sin^3x}$

Solution:

We have,

= $\lim_{x\to0}\frac{log|1+x^3|}{sin^3x}$

= $\lim_{x\to0}\frac{\frac{log|1+x^3|}{x^3}}{\frac{sin^3x}{x^3}}$

= $\lim_{x\to0}\frac{\frac{log|1+x^3|}{x^3}}{\left(\frac{sinx}{x}\right)^3}$

= $\frac{\lim_{x\to0}\frac{log|1+x^3|}{x^3}}{\lim_{x\to0}\left(\frac{sinx}{x}\right)^3}$

We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ and $lim_{x\to0}\frac{sinx}{x}=1$ . So, we get,

= 1

Question 28. $lim_{x\to\frac{π}{2}}\frac{a^{cotx}-a^{cosx}}{cotx-cosx}$

Solution:

We have,

= $lim_{x\to\frac{π}{2}}\left[\frac{a^{cotx}-a^{cosx}}{cotx-cosx}\right]$

= $lim_{x\to\frac{π}{2}}a^{cosx}\left[\frac{a^{cotx-cosx}-1}{cotx-cosx}\right]$

We know, $lim_{x\to0}\frac{log(1+x)}{x}=1$ . So, we have,

= $a^{cos\frac{π}{2}}×loga$

= a⁰ × log a

= log a

Question 29. $lim_{x\to0}\frac{e^x-1}{\sqrt{1-cosx}}$

Solution:

We have,

= $lim_{x\to0}\frac{e^x-1}{\sqrt{1-cosx}}$

= $lim_{x\to0}\frac{(e^x-1)(\sqrt{1+cosx})}{(\sqrt{1-cosx})(\sqrt{1+cosx})}$

= $lim_{x\to0}\frac{(e^x-1)(\sqrt{1+cosx})}{\sqrt{1-cos^2x}}$

= $lim_{x\to0}\frac{(e^x-1)(\sqrt{1+cosx})}{\sqrt{sin^2x}}$

= $lim_{x\to0}\frac{(e^x-1)(\sqrt{1+cosx})}{sinx}$

As numerator and denominator are both zero for x = 0, therefore limit cannot exist.

Question 30. $lim_{x\to0}\frac{e^x-e^5}{x-5}$

Solution:

We have,

= $lim_{x\to0}\frac{e^x-e^5}{x-5}$

Let x = h + 5. We get,

= $lim_{h\to0}\frac{e^{h+5}-e^5}{h+5-5}$

= $lim_{h\to0}\frac{e^{h+5}-e^5}{h}$

= $e^5lim_{h\to0}\frac{e^h-1}{h}$

We know, $lim_{x\to0}\frac{a^x-1}{x}=loga$ . So, we have,

= e⁵ × log e

= e⁵

My Personal Notes

Last Updated : 04 May, 2021

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Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 2

Evaluate the following limits:

Question 16. $\lim_{x\to0}\frac{sin2x}{e^x-1}$

Question 17. $\lim_{x\to0}\frac{e^{sinx}-1}{x}$

Question 18. $\lim_{x\to0}\frac{e^{2x}-e^x}{sin2x}$

Question 19. $\lim_{x\to0}\frac{logx-loga}{x-a}$

Question 20. $\lim_{x\to0}\frac{log(a+x)-log(a-x)}{x}$

Question 21. $\lim_{x\to0}\frac{log(2+x)+log(0.5)}{x}$

Question 22. $\lim_{x\to0}\frac{log(a+x)-loga}{x}$

Question 23. $\lim_{x\to0}\frac{log(3+x)-log(3-x)}{x}$

Question 24. $\lim_{x\to0}\frac{8^x-2^x}{x}$

Question 25. $\lim_{x\to0}\frac{x(2^x-1)}{cosx}$

Question 26. $\lim_{x\to0}\frac{\sqrt{1+x}-1}{log(1+x)}$

Question 27. $\lim_{x\to0}\frac{log|1+x^3|}{sin^3x}$

Question 28. $lim_{x\to\frac{π}{2}}\frac{a^{cotx}-a^{cosx}}{cotx-cosx}$

Question 29. $lim_{x\to0}\frac{e^x-1}{\sqrt{1-cosx}}$

Question 30. $lim_{x\to0}\frac{e^x-e^5}{x-5}$

Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Related Articles

Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 2

Evaluate the following limits:

Question 16.

Question 17.

Question 18.

Question 19.

Question 20.

Question 21.

Question 22.

Question 23.

Question 24.

Question 25.

Question 26.

Question 27.

Question 28.

Question 29.

Question 30.

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Data Structures and Algorithms - Self Paced

CBSE Class 12 Computer Science

GATE CS & IT 2024

Question 16. $\lim_{x\to0}\frac{sin2x}{e^x-1}$

Question 17. $\lim_{x\to0}\frac{e^{sinx}-1}{x}$

Question 18. $\lim_{x\to0}\frac{e^{2x}-e^x}{sin2x}$

Question 19. $\lim_{x\to0}\frac{logx-loga}{x-a}$

Question 20. $\lim_{x\to0}\frac{log(a+x)-log(a-x)}{x}$

Question 21. $\lim_{x\to0}\frac{log(2+x)+log(0.5)}{x}$

Question 22. $\lim_{x\to0}\frac{log(a+x)-loga}{x}$

Question 23. $\lim_{x\to0}\frac{log(3+x)-log(3-x)}{x}$

Question 24. $\lim_{x\to0}\frac{8^x-2^x}{x}$

Question 25. $\lim_{x\to0}\frac{x(2^x-1)}{cosx}$

Question 26. $\lim_{x\to0}\frac{\sqrt{1+x}-1}{log(1+x)}$

Question 27. $\lim_{x\to0}\frac{log|1+x^3|}{sin^3x}$

Question 28. $lim_{x\to\frac{π}{2}}\frac{a^{cotx}-a^{cosx}}{cotx-cosx}$

Question 29. $lim_{x\to0}\frac{e^x-1}{\sqrt{1-cosx}}$

Question 30. $lim_{x\to0}\frac{e^x-e^5}{x-5}$