Differentiation (Part-38) | S N De

 

$\,13.\,$ If $\,\lim_{x \to 0} ~(1+x)^{\frac 1x}=e,\,\,$ show that,

$\,(i)\, \lim_{x \to 0}~~(1+2x)^{\frac 1x}=e^2.$

Sol.  $\, \lim_{x \to 0}~~(1+2x)^{\frac 1x}\\=\lim_{x \to 0}\left[(1+2x)^{\frac{1}{2x}}\right]^2\\=\lim_{z \to 0}\left[(1+z)^{\frac{1}{z}}\right]^2~~[\text{where}~~z=2x]\\=\left[\lim_{z \to 0}~(1+z)^{\frac{1}{z}}\right]^2\\=e^2\,\,\text{(proved)}$

$\,13(ii)~~\lim_{x \to 0}~~(1-3x)^{\frac 3x}=e^{-9}$

Sol. $~~\lim_{x \to 0}~~(1-3x)^{\frac 3x}\\=\lim_{x \to 0}\left[(1-3x)^{\frac{1}{-3x}}\right]^{-9}\\=\left[\lim_{z \to 0}(1+z)^{\frac{1}{z}}\right]^{-9} \\ ~~~[\text{where,}\,\,z=-3x,\,\text{as}\,~~ x \to 0, z \to 0]\\=e^{-9}\,\,\text{(proved)}$

$\,13(iii)\,\lim_{x \to 1} x^{\frac{1}{1-x}}=e^{-1}$

Sol. $\,\,\lim_{x \to 1} x^{\frac{1}{1-x}}\\=\lim_{z \to 0} (1+z)^{-\frac 1z} \\~~~~[\text{where},\,\,z=x-1,\,\,\text{as}\,~~ x \to 1, z \to 0]\\=\left[\lim_{z \to 0} (1+z)^{\frac 1z}\right]^{-1}\\=e^{-1}\,\,\text{(proved)}$

$\,13(iv).\, \lim_{x \to 0} (1+4x)^{\frac{x+2}{x}}=e^8.$

Sol. $\,\,\lim_{x \to 0} (1+4x)^{\frac{x+2}{x}}\\=\lim_{x \to 0}(1+4x)^{1+\frac 2x}\\=\lim_{x \to 0} [(1+4x) \times (1+4x)^{\frac 2x}]\\=\lim_{x \to 0}(1+4x)\cdot  \lim_{x \to 0}(1+4x)^{\frac 2x}\\=1 \cdot \lim_{x \to 0} \left[(1+4x)^{\frac{1}{4x}}\right]^8\\=\left[\lim_{z \to 0} ((1+z)^{\frac 1z})\right]^8~~[\text{where}~~z=4x]\\=e^8~~~\text{(proved)}$

$\,14.\,$ If $\,f(x)=\lim_{n  \to \infty}\frac{1}{1+n\sin^2 \pi x},\,\,$ find $\,f(x)\,$ for all real values of $\,x.$

Sol. If $\,x\,$ is any integer, then $\,\sin(\pi x)=0.$

$\therefore f(x)=\frac{1}{1+n \cdot 0}=1.$

For any  fractional values of $\,x,\,~~\sin(\pi x) \neq 0.$

So, in that case, $(1+n \sin^2 \pi x) \to \infty,~~\text{as}~~ n \to \infty.$           

Hence, for any fractional values of $\,x,$ 

$\,\,f(x)=\lim_{n  \to \infty}\frac{1}{1+n\sin^2 \pi x}=0$

$\,15.\,$ Evaluate : $\,\,\lim_{x \to \frac{\pi}{4}}\frac{4\sqrt2-(\cos x+\sin x)^5}{1-\sin 2x}$

Sol. $\,\,\lim_{x \to \frac{\pi}{4}}\frac{4\sqrt2-(\cos x+\sin x)^5}{1-\sin 2x}\\=\lim_{x \to \frac{\pi}{4}}\frac{4\sqrt 2-\{(\cos x+\sin x)^2\}^\frac 52}{1-\sin 2x}\\=\lim_{x \to \frac{\pi}{4}}\frac{(2)^{\frac 52}-(1+\sin 2x)^{\frac 52}}{2-(1+\sin 2x)}\\=\lim_{z \to 2}\frac{2^{5/2}-z^{5/2}}{2-z}~~~[*]\\=\lim_{z \to 2}\frac{z^{5/2}-2^{5/2}}{z-2}\\=\frac 52 \cdot 2^{5/2-1}\\=\frac 52 \cdot 2^{3/2}\\=5 \sqrt2\,\,\,\text{(ans.)}$

Note[*]: Let $\,1+\sin 2x=z,\, \text{and so as}~~ x \to \frac{\pi}{4}, z \to 2.$ 

$\,16.\,$ Evaluate : $\,\lim_{x \to 4} \frac{(\cos \alpha)^x-(\sin\alpha)^x-\cos 2\alpha}{x-4}~~\left(0 <\alpha<\pi/4\right)$

Sol.  Let $\,L=\lim_{x \to 4} \frac{(\cos \alpha)^x-(\sin\alpha)^x-\cos 2\alpha}{x-4}~~ \rightarrow(1)$

Since the given limit $\,L,\,$ is in $\frac 00\,$ form, so from $\,(1),\,$

we get by applying L'Hospital's rule,

$\,L=\lim_{x \to 4}\frac{\cos^x \alpha. \log_e(\cos\alpha)-\sin^x\alpha. \log_e(\sin\alpha)}{1}\\~~~=\cos^4 \alpha. \log_e(\cos\alpha)-\sin^4\alpha. \log_e(\sin\alpha)$