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} = ex
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)} = eab−0
= eab
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}} = ecot 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.