Question Number 1594 by 123456 last updated on 24/Aug/15
$${f}\left({x}\right)=\underset{\mathrm{0}} {\overset{{x}} {\int}}{t}^{{x}} {dt},{x}>−\mathrm{1} \\ $$$$\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)=? \\ $$$${f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)=? \\ $$
Commented by 112358 last updated on 25/Aug/15
![f(x)=∫_0 ^x t^x dt=(t^(x+1) /(x+1))∣_0 ^x =(1/(x+1))(x^(x+1) −0^(x+1) ) f(x)=(1/(x+1))x^(x+1) ,x>−1 f(x+1)−f(x)=(1/(x+2))(x+1)^(x+2) −(1/(x+1))x^(x+1) rhs=(((x+1)^2 )/(x+2))(x+1)^x −(x/(x+1))x^x =((x^2 +2x+1)/(x+2))(x+1)^x −(x/(x+1))x^x =((x(x+2)+1)/(x+2))(x+1)^x −(x/(x+1))x^x =x[(x+1)^x −(x^x /(x+1))]+(((x+1)^x )/(x+2)) =x[(((x+1)^(x+1) −x^x )/(x+1))]+(((x+1)^x )/(x+2)) =((x{(x+1)^(x+1) −x^x }(x+2)+(x+1)^(x+1) )/((x+1)(x+2))) =((x^2 (x+1)^(x+1) +2x(x+1)^(x+1) +(x+1)^(x+1) −x^(x+2) −2x^(x+1) )/((x+1)(x+2))) f(x+1)−f(x)=(((x+1)^(x+3) −x^(x+1) (x+2))/((x+1)(x+2))) f(x)=(1/(1+x))x^(x+1) =(x/(1+x))x^x =(1−(1/(x+1)))x^x Let L=lim_(x→−1^+ ) f(x) (if this limit exists) L=lim_(x→−1^+ ) f(x)=lim_(x→−1^+ ) [(1−(1/(x+1)))x^x ] L=[lim_(x→−1^+ ) (1−(1/(x+1)))]×[lim_(x→−1^+ ) x^x ] Let p=lim_(x→−1^+ ) x^x p=lim_(x→−1^+ ) x^x =lim_(x→−1^+ ) e^(lnx^x ) =lim_(x→−1^+ ) e^(xlnx) p=e^(lim_(x→−1^+ ) xlnx) =e^((lim_(x→−1^+ ) x)(lim_(x→−1^+ ) lnx)) p=e^(−1×ln(−1)) =e^(ln(−1)) =e^(ln1+iπ) p=e^(iπ) =−1 Let q=lim_(x→−1^+ ) (1−(1/(x+1))) (1/(1+x))=(1+x)^(−1) ⇒ (1/(1+x))=e^(−ln(x+1)) ∴q=lim_(x→−1^+ ) 1−exp(−lim_(x→−1^+ ) ln(x+1)) Now, lim_(x→−1^+ ) ln(x+1)=−∞ ∴ q=1−exp(−1×−∞)=−∞ ∵ L=p×q⇒L=−∞×−1=∞ L does not exist. ⇒lim_(x→−1^+ ) f(x) does not exist](https://www.tinkutara.com/question/Q1595.png)
$${f}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {t}^{{x}} {dt}=\frac{{t}^{{x}+\mathrm{1}} }{{x}+\mathrm{1}}\mid_{\mathrm{0}} ^{{x}} =\frac{\mathrm{1}}{{x}+\mathrm{1}}\left({x}^{{x}+\mathrm{1}} −\mathrm{0}^{{x}+\mathrm{1}} \right) \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}+\mathrm{1}}{x}^{{x}+\mathrm{1}} \:,{x}>−\mathrm{1} \\ $$$${f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)=\frac{\mathrm{1}}{{x}+\mathrm{2}}\left({x}+\mathrm{1}\right)^{{x}+\mathrm{2}} −\frac{\mathrm{1}}{{x}+\mathrm{1}}{x}^{{x}+\mathrm{1}} \\ $$$${rhs}=\frac{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{{x}+\mathrm{2}}\left({x}+\mathrm{1}\right)^{{x}} −\frac{{x}}{{x}+\mathrm{1}}{x}^{{x}} \: \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}}{{x}+\mathrm{2}}\left({x}+\mathrm{1}\right)^{{x}} −\frac{{x}}{{x}+\mathrm{1}}{x}^{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:=\frac{{x}\left({x}+\mathrm{2}\right)+\mathrm{1}}{{x}+\mathrm{2}}\left({x}+\mathrm{1}\right)^{{x}} −\frac{{x}}{{x}+\mathrm{1}}{x}^{{x}} \\ $$$$\:\:\:\:\:\:\:\:\:={x}\left[\left({x}+\mathrm{1}\right)^{{x}} −\frac{{x}^{{x}} }{{x}+\mathrm{1}}\right]+\frac{\left({x}+\mathrm{1}\right)^{{x}} }{{x}+\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:={x}\left[\frac{\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} −{x}^{{x}} }{{x}+\mathrm{1}}\right]+\frac{\left({x}+\mathrm{1}\right)^{{x}} }{{x}+\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\frac{{x}\left\{\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} −{x}^{{x}} \right\}\left({x}+\mathrm{2}\right)+\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} }{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\frac{{x}^{\mathrm{2}} \left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} +\mathrm{2}{x}\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} +\left({x}+\mathrm{1}\right)^{{x}+\mathrm{1}} −{x}^{{x}+\mathrm{2}} −\mathrm{2}{x}^{{x}+\mathrm{1}} }{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$${f}\left({x}+\mathrm{1}\right)−{f}\left({x}\right)=\frac{\left({x}+\mathrm{1}\right)^{{x}+\mathrm{3}} −{x}^{{x}+\mathrm{1}} \left({x}+\mathrm{2}\right)}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)} \\ $$$${f}\left({x}\right)=\frac{\mathrm{1}}{\mathrm{1}+{x}}{x}^{{x}+\mathrm{1}} =\frac{{x}}{\mathrm{1}+{x}}{x}^{{x}} =\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){x}^{{x}} \\ $$$${Let}\:{L}=\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)\:\left({if}\:{this}\:{limit}\:{exists}\right) \\ $$$${L}=\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)=\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}\left[\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right){x}^{{x}} \right] \\ $$$${L}=\left[\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right)\right]×\left[\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{x}^{{x}} \right] \\ $$$${Let}\:{p}=\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{x}^{{x}} \\ $$$${p}=\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{x}^{{x}} =\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{e}^{{lnx}^{{x}} } =\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{e}^{{xlnx}} \\ $$$${p}={e}^{\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{xlnx}} ={e}^{\left(\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{x}\right)\left(\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{lnx}\right)} \\ $$$${p}={e}^{−\mathrm{1}×{ln}\left(−\mathrm{1}\right)} ={e}^{{ln}\left(−\mathrm{1}\right)} ={e}^{{ln}\mathrm{1}+{i}\pi} \\ $$$${p}={e}^{{i}\pi} =−\mathrm{1} \\ $$$${Let}\:{q}=\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}\left(\mathrm{1}−\frac{\mathrm{1}}{{x}+\mathrm{1}}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{1}+{x}}=\left(\mathrm{1}+{x}\right)^{−\mathrm{1}} \Rightarrow\:\frac{\mathrm{1}}{\mathrm{1}+{x}}={e}^{−{ln}\left({x}+\mathrm{1}\right)} \\ $$$$\therefore{q}=\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}1}−{exp}\left(−\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{ln}\left({x}+\mathrm{1}\right)\right) \\ $$$${Now},\:\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{ln}\left({x}+\mathrm{1}\right)=−\infty \\ $$$$\therefore\:{q}=\mathrm{1}−{exp}\left(−\mathrm{1}×−\infty\right)=−\infty \\ $$$$\because\:{L}={p}×{q}\Rightarrow{L}=−\infty×−\mathrm{1}=\infty \\ $$$${L}\:{does}\:{not}\:{exist}. \\ $$$$\Rightarrow\underset{{x}\rightarrow−\mathrm{1}^{+} } {\mathrm{lim}}{f}\left({x}\right)\:{does}\:{not}\:{exist} \\ $$$$ \\ $$
Commented by Rasheed Soomro last updated on 24/Aug/15
$${Excelent}! \\ $$