Answer to Question #134532 in Calculus for sire

Question

using L’Hˆopital’s in the following questions

1) I = limx-0 [1/x - 2/ln(1 + 2x)]
2) limx-0 [e^x + x]^1/x
3) limx-1+ (ln x)(tan(pi*x/2))

Answer 1

$\lim\limits_{x\rightarrow0}(\frac{1}{x} -\frac{2}{ln(1+2x)})=\lim\limits_{x\rightarrow0}(\frac{ln(1+2x)-2x}{xln(1+2x)})=\lim\limits_{x\rightarrow0}\frac{0}{0}$ $=\lim\limits_{x\rightarrow0}(\frac{(ln(1+2x)-2x)'}{(xln(1+2x))'})=$

$\lim\limits_{x\rightarrow0}(\frac{(ln(1+2x)-2x)'}{(xln(1+2x))'})=\lim\limits_{x\rightarrow0}[\frac{\frac{2}{1+2x}-2}{ln(1+2x)+\frac{2x}{1+2x}}]= \lim\limits_{x\rightarrow0}\frac{-4x}{(1+2x)ln(1+2x)+2x}=$

$\lim\limits_{x\rightarrow0}\frac{0}{0}=$ $\lim\limits_{x\rightarrow0}\frac{-(4x)'}{((1+2x)ln(1+2x)+2x)'}=$ $\lim\limits_{x\rightarrow0}\frac{-4}{2ln(1+2x)+(1+2x)*\frac{2}{1+2x}+2}=\lim\limits_{x\rightarrow0}\frac{-4}{2ln(1+2x)+4}=-1$

$\lim\limits_{x\rightarrow0}(e^x+x)^{\frac{1}{x}}=\lim\limits_{x\rightarrow0}e^{ln(e^x+x)^\frac{1}{x}}$ -we use equality a= e^lna

$\lim\limits_{x\rightarrow0}ln(e^x+x)^\frac{1}{x} = \lim\limits_{x\rightarrow0}\frac{ln(e^x+x)}{x} =$ $\lim\limits_{x\rightarrow0}\frac{0}{0}$-we use equality log(a^b) = blog(a)

$\lim\limits_{x\rightarrow0}\frac{ln(e^x+x)}{x} = \lim\limits_{x\rightarrow0}(ln(e^x+x)^\frac{1}{x} = \lim\limits_{x\rightarrow0}\frac{(ln(e^x+x))'}{x'} =$ $\lim\limits_{x\rightarrow0}\frac{\frac{e^x+1}{e^x+x}}{1}=2$ hence

$\lim\limits_{x\rightarrow0}(e^x+x)^{\frac{1}{x}}=e^2$

$\lim\limits_{x\rightarrow1+}lnx*tan(\frac{\pi*x}{2})=\lim\limits_{x\rightarrow1+}\frac{lnx}{cot(\frac{\pi*x}{2})}= \lim\limits_{x\rightarrow1+}\frac{0}{0}$ we use equality tan(x) = 1/cot(x)

$\lim\limits_{x\rightarrow1+}\frac{lnx}{cot(\frac{\pi*x}{2})} =\lim\limits_{x\rightarrow1+}\frac{(lnx)'}{(cot(\frac{\pi*x}{2}))'}=\lim\limits_{x\rightarrow1+}\frac{\frac{1}{x}}{-\frac{\pi}{2}*csc^2(\frac{\pi*x}{2})}=$ $-\frac{2}{\pi}$

Answer:

$\lim\limits_{x\rightarrow0}(\frac{1}{x} -\frac{2}{ln(1+2x)})=-1$

$\lim\limits_{x\rightarrow0}(e^x+x)^{\frac{1}{x}}=e^2$

$\lim\limits_{x\rightarrow1+}lnx*tan(\frac{\pi*x}{2})=-\frac{2}{\pi}$