Class 11 RD Sharma Solutions - Chapter 29 Limits - Exercise 29.11

Evaluate the following limits:

Question 1. \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n

Solution:

We know if \lim_{x\to{a}}f(x)=\lim_{x\to{a}}g(x)=0   such that \lim_{x\to{a}}\frac{f(x)}{g(x)}   exists, then \lim_{x\to{a}}(1+f(x))^\frac{1}{g(x)}=e^{\lim_{x\to{a}}\frac{f(x)}{g(x)}}

We have,

= \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n

= e^{\lim_{n\to\infty}(\frac{x}{n})n}

= e^x

Question 2. \lim_{x\to0^+}\left(1+tan^2\sqrt{x}\right)^\frac{1}{2x}

Solution:

We have,

= \lim_{x\to0^+}\left(1+tan^2\sqrt{x}\right)^\frac{1}{2x}

= e^{\lim_{x\to0^+}(\frac{tan^2\sqrt{x}}{2x})}

= e^{\lim_{x\to0^+}\left(\frac{sin^2\sqrt{x}}{2xcos^2\sqrt{x}}\right)}

= e^{\lim_{x\to0^+}\frac{1}{2}\left(\frac{sin\sqrt{x}}{\sqrt{x}}\right)^2\lim_{x\to0^+}\left(\frac{1}{cos^2\sqrt{x}}\right)}

= e^{\frac{1}{2}}

= \sqrt{e}

Question 3. \lim_{x\to0}\left(cosx\right)^\frac{1}{sinx}

Solution:

We have,

= \lim_{x\to0}\left(cosx\right)^\frac{1}{sinx}

= \lim_{x\to0}\left(1+cosx-1\right)^\frac{1}{sinx}

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

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

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

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

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

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

= e^{\lim_{x\to0}\left(-tanx\right)}

= e⁰

= 1

Question 4. \lim_{x\to0}\left(cosx+sinx\right)^\frac{1}{x}

Solution:

We have,

= \lim_{x\to0}\left(cosx+sinx\right)^\frac{1}{x}

= \lim_{x\to0}\left(1+(cosx+sinx-1)\right)^\frac{1}{x}

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

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

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

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

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

= e¹⁻⁰

= e

Question 5. \lim_{x\to0}\left(cosx+asinbx\right)^\frac{1}{x}  

Solution:

We have,

= \lim_{x\to0}\left(cosx+asinbx\right)^\frac{1}{x}

= \lim_{x\to0}\left(1+(cosx+asinbx-1)\right)^\frac{1}{x}

= e^{\lim_{x\to0}\left(\frac{cosx+asinbx-1}{x}\right)}

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

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

= e^{\lim_{x\to0}\left(\frac{absinx}{bx}\right)-\lim_{x\to0}\left(\frac{2sin^2\frac{x}{2}}{2(\frac{x}{2})}\right)}

= e^{\lim_{x\to0}\left(\frac{absinx}{bx}\right)-\lim_{x\to0}\left(\frac{sin\frac{x}{2}sin\frac{x}{2}}{\frac{x}{2}}\right)}

= e^ab−0

= e^ab

Question 6. \lim_{x\to\infty}\left(\frac{x^2+2x+3}{2x^2+x+5}\right)^{\frac{3x-2}{3x+2}}

Solution:

We have,

= \lim_{x\to\infty}\left(\frac{x^2+2x+3}{2x^2+x+5}\right)^{\frac{3x-2}{3x+2}}

= e^{\lim_{x\to\infty}\left[\left(\frac{3x-2}{3x+2}\right)ln\left(\frac{x^2+2x+3}{2x^2+x+5}\right)\right]}

= e^{\lim_{x\to\infty}\left[\left(\frac{3-\frac{2}{x}}{3+\frac{2}{x}}\right)ln\left(\frac{1+\frac{2}{x}+\frac{3}{x^2}}{2+\frac{1}{x}+\frac{5}{x^2}}\right)\right]}

= e^{1.ln(\frac{1}{2})}

= \frac{1}{2}

Question 7. \lim_{x\to1}\left(\frac{x^3+2x^2+x+1}{x^2+2x+3}\right)^{\frac{1-cos(x-1)}{(x-1)^2}}

Solution:

We have,

= \lim_{x\to1}\left(\frac{x^3+2x^2+x+1}{x^2+2x+3}\right)^{\frac{1-cos(x-1)}{(x-1)^2}}

= e^{\lim_{x\to1}\left[{\frac{1-cos(x-1)}{(x-1)^2}}ln\left(\frac{x^3+2x^2+x+1}{x^2+2x+3}\right)\right]}

= e^{\lim_{x\to1}\left[{\frac{2sin^2(\frac{x-1}{2})}{4\left(\frac{x-1}{2}\right)^2}}ln\left(\frac{x^3+2x^2+x+1}{x^2+2x+3}\right)\right]}

= e^{\frac{2}{4}ln(\frac{5}{6})}

= e^{ln(\frac{5}{6})^{\frac{1}{2}}}

= (\frac{5}{6})^{\frac{1}{2}}

Question 8.  \lim_{x\to0}\left[\frac{e^x+e^{-x}-2}{x^2}\right]^{\frac{1}{x^2}}

Solution:

We have,

= \lim_{x\to0}\left[\frac{e^x+e^{-x}-2}{x^2}\right]^{\frac{1}{x^2}}  

= e^{\lim_{x\to0}\left[\frac{\frac{e^x+e^{-x}-2}{x^2}}{x^2}\right]}

Applying L'Hospital's Rule, we get,

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

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

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

= e^{\frac{1}{2}\lim_{x\to0}\left(\frac{1+x}{6+6x+x^2}\right)}

= e^{\frac{1}{12}}

Question 9. \lim_{x\to{a}}\left[\frac{sinx}{sina}\right]^{\frac{1}{x-a}}

Solution:

We have,

= \lim_{x\to{a}}\left[\frac{sinx}{sina}\right]^{\frac{1}{x-a}}

= \lim_{x\to{a}}\left[1+(\frac{sinx}{sina}-1)\right]^{\frac{1}{x-a}}

= e^{\lim_{x\to{a}}\left(\frac{(\frac{sinx}{sina}-1)}{x-a}\right)}

= e^{\lim_{x\to{a}}\frac{(\frac{sinx-sina}{sina})}{x-a}}

= e^{\lim_{x\to{a}}\left(\frac{sinx-sina}{sina(x-a)}\right)}

= e^{\lim_{x\to{a}}\left(\frac{2cos(\frac{x+a}{2})sin(\frac{x-a}{2})}{sina(x-a)}\right)}

= e^{\lim_{x\to{a}}\left(\frac{2cos(\frac{x+a}{2})}{sina}\right)lim_{x\to{a}}\left(\frac{sin(\frac{x-a}{2})}{2(\frac{x-a}{2})}\right)}

= e^{\frac{2cosa}{2sina}}

= e^{cot a}

Question 10. \lim_{x\to\infty}\left(\frac{3x^2+1}{4x^2-1}\right)^{\frac{x^3}{1+x}}

Solution:

We have,

= \lim_{x\to\infty}\left(\frac{3x^2+1}{4x^2-1}\right)^{\frac{x^3}{1+x}}

= \lim_{x\to\infty}\left(1+\frac{-x^2+2}{4x^2-1}\right)^{\frac{x^3}{1+x}}

= e^{\lim_{x\to\infty}\left[(\frac{-x^2+2}{4x^2-1}){(\frac{x^3}{1+x}})\right]}

= e^{\lim_{x\to\infty}\left(\frac{-x^5+2x^3}{4x^2-1+4x^3-x}\right)}

= e^−∞

= 1/e^∞

= 0

Summary

Exercise 29.11 covers evaluating limits of various functions, including algebraic, trigonometric, and exponential functions. Students learn to apply limit rules and techniques, such as the product rule, quotient rule, and chain rule. Limits measure the behavior of functions as the input changes. Understanding limits is crucial for calculus and its applications. Practice questions reinforce learning and application. Limits help model real-world phenomena and make predictions.