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}

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}}

Aslim_{x\to0}\frac{a^x-1}{x}=loga andlim_{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)

Aslim_{x\to0}\frac{a^x-1}{x}=loga andlim_{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)

Aslim_{x\to0}\frac{a^x-1}{x}=loga andlim_{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)

Aslim_{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)

Aslim_{x\to0}\frac{a^x-1}{x}=loga andlim_{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 andlim_{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