第11类RD Sharma解决方案–第29章限制–练习29.11

评估以下限制：

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

解决方案：

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

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

解决方案：

We have,

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

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

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

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

= $e^{\frac{1}{2}}$

= $\sqrt{e}$

问题3。 $\lim_{x\to0}\left(cosx\right)^\frac{1}{sinx}$

解决方案：

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

问题4。 $\lim_{x\to0}\left(cosx+sinx\right)^\frac{1}{x}$

解决方案：

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

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

解决方案：

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

问题6。 $\lim_{x\to\infty}\left(\frac{x^2+2x+3}{2x^2+x+5}\right)^{\frac{3x-2}{3x+2}}$

解决方案：

We have,

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

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

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

= $e^{1.ln(\frac{1}{2})}$

= $\frac{1}{2}$

问题7。 $\lim_{x\to1}\left(\frac{x^3+2x^2+x+1}{x^2+2x+3}\right)^{\frac{1-cos(x-1)}{(x-1)^2}}$

解决方案：

We have,

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

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

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

= $e^{\frac{2}{4}ln(\frac{5}{6})}$

= $e^{ln(\frac{5}{6})^{\frac{1}{2}}}$

= $(\frac{5}{6})^{\frac{1}{2}}$

问题8。 $\lim_{x\to0}\left[\frac{e^x+e^{-x}-2}{x^2}\right]^{\frac{1}{x^2}}$

解决方案：

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,

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

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

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

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

= $e^{\frac{1}{12}}$

问题9。 $\lim_{x\to{a}}\left[\frac{sinx}{sina}\right]^{\frac{1}{x-a}}$

解决方案：

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}

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

解决方案：

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