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Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 1
  • Last Updated : 04 May, 2021

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}}

Aslim_{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}



Aslim_{x\to0}\frac{a^x-1}{x}=loga , we get,

=(log_ea)^2×\frac{1}{a^0}

= (loge 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}

Aslim_{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}

Aslim_{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

Aslim_{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

Aslim_{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)

Aslim_{x\to0}\frac{a^x-1}{x}=loga , we get,

= m log a − n log b

= log am − 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}

Aslim_{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}

Aslim_{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}

Aslim_{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}}

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



Aslim_{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}

Aslim_{x\to0}\frac{a^x-1}{x}=loga andlim_{x\to0}\frac{sinx}{x}=1 , we get,

= log e + 1

= 1 + 1

= 2




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