f(x)=∫_0 ^x t^x dt,x>−1 lim_(x→−1^+ ) f(x)=? f(x+1)−f(x)=?


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Question Number 1594 by 123456 last updated on 24/Aug/15

$f (x) = \underset{0}{\int^{x}} t^{x} d t, x > - 1$ $\lim_{x \to - 1^{+}} f (x) = ?$ $f (x + 1) - f (x) = ?$
Commented by112358 last updated on 25/Aug/15
$f(x)=∫_0 ^x t^x dt=(t^(x+1) /(x+1))∣_0 ^x =(1/(x+1))(x^(x+1) −0^(x+1) ) f(x)=(1/(x+1))x^(x+1) ,x>−1 f(x+1)−f(x)=(1/(x+2))(x+1)^(x+2) −(1/(x+1))x^(x+1) rhs=(((x+1)^2 )/(x+2))(x+1)^x −(x/(x+1))x^x =((x^2 +2x+1)/(x+2))(x+1)^x −(x/(x+1))x^x =((x(x+2)+1)/(x+2))(x+1)^x −(x/(x+1))x^x =x[(x+1)^x −(x^x /(x+1))]+(((x+1)^x )/(x+2)) =x[(((x+1)^(x+1) −x^x )/(x+1))]+(((x+1)^x )/(x+2)) =((x{(x+1)^(x+1) −x^x }(x+2)+(x+1)^(x+1) )/((x+1)(x+2))) =((x^2 (x+1)^(x+1) +2x(x+1)^(x+1) +(x+1)^(x+1) −x^(x+2) −2x^(x+1) )/((x+1)(x+2))) f(x+1)−f(x)=(((x+1)^(x+3) −x^(x+1) (x+2))/((x+1)(x+2))) f(x)=(1/(1+x))x^(x+1) =(x/(1+x))x^x =(1−(1/(x+1)))x^x Let L=lim_(x→−1^+ ) f(x) (if this limit exists) L=lim_(x→−1^+ ) f(x)=lim_(x→−1^+ ) [(1−(1/(x+1)))x^x ] L=[lim_(x→−1^+ ) (1−(1/(x+1)))]×[lim_(x→−1^+ ) x^x ] Let p=lim_(x→−1^+ ) x^x p=lim_(x→−1^+ ) x^x =lim_(x→−1^+ ) e^(lnx^x ) =lim_(x→−1^+ ) e^(xlnx) p=e^(lim_(x→−1^+ ) xlnx) =e^((lim_(x→−1^+ ) x)(lim_(x→−1^+ ) lnx)) p=e^(−1×ln(−1)) =e^(ln(−1)) =e^(ln1+iπ) p=e^(iπ) =−1 Let q=lim_(x→−1^+ ) (1−(1/(x+1))) (1/(1+x))=(1+x)^(−1) ⇒ (1/(1+x))=e^(−ln(x+1)) ∴q=lim_(x→−1^+ ) 1−exp(−lim_(x→−1^+ ) ln(x+1)) Now, lim_(x→−1^+ ) ln(x+1)=−∞ ∴ q=1−exp(−1×−∞)=−∞ ∵ L=p×q⇒L=−∞×−1=∞ L does not exist. ⇒lim_(x→−1^+ ) f(x) does not exist$
$f (x) = \int_{0}^{x} t^{x} d t = \frac{t^{x + 1}}{x + 1} ∣_{0}^{x} = \frac{1}{x + 1} (x^{x + 1} - 0^{x + 1})$ $f (x) = \frac{1}{x + 1} x^{x + 1}, x > - 1$ $f (x + 1) - f (x) = \frac{1}{x + 2} {(x + 1)}^{x + 2} - \frac{1}{x + 1} x^{x + 1}$ $r h s = \frac{{(x + 1)}^{2}}{x + 2} {(x + 1)}^{x} - \frac{x}{x + 1} x^{x}$ $= \frac{x^{2} + 2 x + 1}{x + 2} {(x + 1)}^{x} - \frac{x}{x + 1} x^{x}$ $= \frac{x (x + 2) + 1}{x + 2} {(x + 1)}^{x} - \frac{x}{x + 1} x^{x}$ $= x [{(x + 1)}^{x} - \frac{x^{x}}{x + 1}] + \frac{{(x + 1)}^{x}}{x + 2}$ $= x [\frac{{(x + 1)}^{x + 1} - x^{x}}{x + 1}] + \frac{{(x + 1)}^{x}}{x + 2}$ $= \frac{x {{(x + 1)}^{x + 1} - x^{x}} (x + 2) + {(x + 1)}^{x + 1}}{(x + 1) (x + 2)}$ $= \frac{x^{2} {(x + 1)}^{x + 1} + 2 x {(x + 1)}^{x + 1} + {(x + 1)}^{x + 1} - x^{x + 2} - 2 x^{x + 1}}{(x + 1) (x + 2)}$ $f (x + 1) - f (x) = \frac{{(x + 1)}^{x + 3} - x^{x + 1} (x + 2)}{(x + 1) (x + 2)}$ $f (x) = \frac{1}{1 + x} x^{x + 1} = \frac{x}{1 + x} x^{x} = (1 - \frac{1}{x + 1}) x^{x}$ $L e t L = \lim_{x \to - 1^{+}} f (x) (i f t h i s l i m i t e x i s t s)$ $L = \lim_{x \to - 1^{+}} f (x) = \lim_{x \to - 1^{+}} [(1 - \frac{1}{x + 1}) x^{x}]$ $L = [\lim_{x \to - 1^{+}} (1 - \frac{1}{x + 1})] \times [\lim_{x \to - 1^{+}} x^{x}]$ $L e t p = \lim_{x \to - 1^{+}} x^{x}$ $p = \lim_{x \to - 1^{+}} x^{x} = \lim_{x \to - 1^{+}} e^{{l n x}^{x}} = \lim_{x \to - 1^{+}} e^{x l n x}$ $p = e^{\lim_{x \to - 1^{+}} x l n x} = e^{(\lim_{x \to - 1^{+}} x) (\lim_{x \to - 1^{+}} l n x)}$ $p = e^{- 1 \times l n (- 1)} = e^{l n (- 1)} = e^{l n 1 + i π}$ $p = e^{i π} = - 1$ $L e t q = \lim_{x \to - 1^{+}} (1 - \frac{1}{x + 1})$ $\frac{1}{1 + x} = {(1 + x)}^{- 1} \Rightarrow \frac{1}{1 + x} = e^{- l n (x + 1)}$ $∴ q = \underset{x \to - 1^{+}}{\lim 1} - e x p (- \lim_{x \to - 1^{+}} l n (x + 1))$ $N o w, \lim_{x \to - 1^{+}} l n (x + 1) = - \infty$ $∴ q = 1 - e x p (- 1 \times - \infty) = - \infty$ $∵ L = p \times q \Rightarrow L = - \infty \times - 1 = \infty$ $L d o e s n o t e x i s t .$ $\Rightarrow \lim_{x \to - 1^{+}} f (x) d o e s n o t e x i s t$
Commented byRasheed Soomro last updated on 24/Aug/15

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