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Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.10 | Set 3
  • Last Updated : 04 May, 2021

Evaluate the following limits:

Question 31.\lim_{x\to0}\frac{e^{x+2}-e^x}{x}

Solution:

We have,

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

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

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



= e2 × log e

= e2

Question 32.\lim_{x\to\frac{π}{2}}\frac{e^{cosx}-1}{cosx}

Solution:

We have,

=\lim_{x\to\frac{π}{2}}\frac{e^{cosx}-1}{cosx}

Let x − π/2 = h. So, we get

=\lim_{h\to0}\frac{e^{cos(\frac{π}{2}+h)}-1}{cos(\frac{π}{2}+h)}

=\lim_{sinh\to0}\frac{e^{-sinh}-1}{-sinh}



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

= log e

= 1

Question 33.\lim_{x\to0}\frac{e^{3+x}-sinx-e^3}{x}

Solution:

We have,

=\lim_{x\to0}\frac{e^{3+x}-sinx-e^3}{x}

=\lim_{x\to0}\frac{e^{3+x}-e^3}{x}-\lim_{x\to0}\frac{sinx}{x}

=e^3\lim_{x\to0}\frac{e^x-1}{x}-\lim_{x\to0}\frac{sinx}{x}

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

= e3 log e − 1



= e3 − 1

Question 34.lim_{x\to0}\frac{e^x-x-1}{x}

Solution:

We have,

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

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

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

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

= log e − 1

= 1 − 1

= 0

Question 35.lim_{x\to0}\frac{e^{3x}-e^{2x}}{x}

Solution:

We have,

=lim_{x\to0}\frac{e^{3x}-e^{2x}}{x}

=lim_{x\to0}\frac{e^{3x}-1-(e^{2x}-1)}{x}

=lim_{x\to0}\frac{e^{3x}-1}{x}-lim_{x\to0}\frac{e^{2x}-1}{x}

=lim_{x\to0}3×(\frac{e^{3x}-1}{3x})-lim_{x\to0}2×(\frac{e^{2x}-1}{2x})

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

= 3 log e − 2 log e

= 3 − 2

= 1

Question 36.lim_{x\to0}\frac{e^{tanx}-1}{tanx}

Solution:

We have,

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

=lim_{tanx\to0}\frac{e^{tanx}-1}{tanx}

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

= log e

= 1

Question 37.lim_{x\to0}\frac{e^{bx}-e^{ax}}{x} , 0 < a < b

Solution:

We have,

=lim_{x\to0}\frac{e^{bx}-e^{ax}}{x}

=lim_{x\to0}\frac{e^{bx}-1-(e^{ax}-1)}{x}

=lim_{x\to0}\frac{e^{bx}-1}{x}-lim_{x\to0}\frac{e^{ax}-1}{x}

=lim_{x\to0}b×\frac{e^{bx}-1}{bx}-lim_{x\to0}a×\frac{e^{ax}-1}{ax}

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

= b log e − a log e

= b − a

Question 38.lim_{x\to0}\frac{e^{tan}-1}{x}

Solution:

We have,

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

=lim_{x\to0}\frac{e^{tan}-1}{tanx}×lim_{x\to0}\frac{tanx}{x}

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

= log e × 1

= 1

Question 39.lim_{x\to0}\frac{e^x-e^{sinx}}{x-sinx}

Solution:

We have,

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

=lim_{x\to0}e^{sinx}\left[\frac{e^{x-sinx}-1}{x-sinx}\right]

=lim_{x\to0}e^{sinx}×lim_{x\to0}\left[\frac{e^{x-sinx}-1}{x-sinx}\right]

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

=lim_{x\to0}e^{sinx}×loge

= e0

= 1

Question 40.lim_{x\to0}\frac{3^{2+x}-9}{x}

Solution:

We have,

=lim_{x\to0}\frac{3^{2+x}-9}{x}

=lim_{x\to0}\frac{3^{2+x}-3^2}{x}

=3^2×lim_{x\to0}(\frac{3^x-1}{x})

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

= 9 × log 3

= 9 log 3

Question 41.lim_{x\to0}\frac{a^x-a^{-x}}{x}

Solution:

We are given,

=lim_{x\to0}\frac{a^x-a^{-x}}{x}

=lim_{x\to0}\frac{a^{2x}-1}{xa^x}

=lim_{x\to0}\frac{a^{2x}-1}{x}×lim_{x\to0}\frac{1}{a^x}

=lim_{x\to0}\frac{2(a^{2x}-1)}{2x}×lim_{x\to0}\frac{1}{a^x}

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

=2loga×\frac{1}{a^0}

= 2 log a

Question 42.lim_{x\to0}\frac{x(e^x-1)}{1-cosx}

Solution:

We have,

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

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

=lim_{x\to0}\frac{e^x-1}{2x}×lim_{x\to0}\frac{4}{\left(\frac{sin\frac{x}{2}}{\frac{x}{2}}\right)^2}

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

=\frac{loge}{2}×4

= 2

Question 43.lim_{x\to\frac{π}{2}}\frac{2^{-cosx}-1}{xsin(x-\frac{π}{2})}

Solution:

We have,

=lim_{x\to\frac{π}{2}}\frac{2^{-cosx}-1}{xsin(x-\frac{π}{2})}

=lim_{x\to\frac{π}{2}}\frac{2^{-sin(\frac{π}{2}-x)}-1}{xsin(x-\frac{π}{2})}

=lim_{x\to\frac{π}{2}}\frac{2^{sin(x-\frac{π}{2})}-1}{xsin(x-\frac{π}{2})}

=lim_{x\to\frac{π}{2}}\frac{2^{sin(x-\frac{π}{2})}-1}{sin(x-\frac{π}{2})}×lim_{x\to\frac{π}{2}}\frac{1}{x}

Let x− π/2 = h in first part. So, we get,

=lim_{h\to0}\frac{2^{sinh}-1}{sinh}×lim_{x\to\frac{π}{2}}\frac{1}{x}

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

=log 2 ×\frac{2}{π}

=\frac{2log2}{π}

=\frac{log4}{π}

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