f-x-0-x-t-x-dt-x-gt-1-lim-x-1-f-x-f-x-1-f-x- Tinku Tara June 3, 2023 Others 0 Comments FacebookTweetPin Question Number 1594 by 123456 last updated on 24/Aug/15 f(x)=∫x0txdt,x>−1limx→−1+f(x)=?f(x+1)−f(x)=? Commented by 112358 last updated on 25/Aug/15 f(x)=∫0xtxdt=tx+1x+1∣0x=1x+1(xx+1−0x+1)f(x)=1x+1xx+1,x>−1f(x+1)−f(x)=1x+2(x+1)x+2−1x+1xx+1rhs=(x+1)2x+2(x+1)x−xx+1xx=x2+2x+1x+2(x+1)x−xx+1xx=x(x+2)+1x+2(x+1)x−xx+1xx=x[(x+1)x−xxx+1]+(x+1)xx+2=x[(x+1)x+1−xxx+1]+(x+1)xx+2=x{(x+1)x+1−xx}(x+2)+(x+1)x+1(x+1)(x+2)=x2(x+1)x+1+2x(x+1)x+1+(x+1)x+1−xx+2−2xx+1(x+1)(x+2)f(x+1)−f(x)=(x+1)x+3−xx+1(x+2)(x+1)(x+2)f(x)=11+xxx+1=x1+xxx=(1−1x+1)xxLetL=limx→−1+f(x)(ifthislimitexists)L=limx→−1+f(x)=limx→−1+[(1−1x+1)xx]L=[limx→−1+(1−1x+1)]×[limx→−1+xx]Letp=limx→−1+xxp=limx→−1+xx=limx→−1+elnxx=limx→−1+exlnxp=elimx→−1+xlnx=e(limx→−1+x)(limx→−1+lnx)p=e−1×ln(−1)=eln(−1)=eln1+iπp=eiπ=−1Letq=limx→−1+(1−1x+1)11+x=(1+x)−1⇒11+x=e−ln(x+1)∴q=lim1x→−1+−exp(−limx→−1+ln(x+1))Now,limx→−1+ln(x+1)=−∞∴q=1−exp(−1×−∞)=−∞∵L=p×q⇒L=−∞×−1=∞Ldoesnotexist.⇒limx→−1+f(x)doesnotexist Commented by Rasheed Soomro last updated on 24/Aug/15 Excelent! Terms of Service Privacy Policy Contact: info@tinkutara.com FacebookTweetPin Post navigation Previous Previous post: Let-is-cube-root-of-unity-and-x-y-z-Z-If-x-y-z-0-prove-that-3-x-y-z-Show-by-an-example-that-the-converse-is-not-true-Next Next post: f-x-f-x-dx-ln-sec-x-c-f-x- Leave a Reply Cancel replyYour email address will not be published. Required fields are marked *Comment * Name * Save my name, email, and website in this browser for the next time I comment.
![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)
