Answer to Question #203006 in Calculus for Femi

Question #203006

Limx→∞ -(x+1)(e1/(x+1) -1)


1
Expert's answer
2021-06-07T03:33:11-0400

We need to calculate lim(x)(x+1)(e1(x+1)1)lim(x\to\infty) -(x+1)(e^\frac {1} {(x+1)} -1)


=lim(x)(1)e1(x+1)11(x+1)=lim(x\to\infty)(-1)\frac {e^\frac {1} {(x+1)} -1 } {\frac {1} {(x+1)}}

It is in 00\frac 0 0 form after putting \infty in place of xx .So L hospital's rule is applicable here.


=(1)lim(x)ddx(e1(x+1)1)ddx1(x+1)(-1) lim(x\to\infty)\frac {\frac {d} {dx}(e^\frac {1} {(x+1)} -1)} {\frac {d} {dx}\frac {1} {(x+1)}}


=(1)lim(x)e1(x+1).1(x+1)21(x+1)2=(-1)lim(x\to\infty)\frac {e^\frac {1} {(x+1)}. \frac {-1} {(x+1)^2}} {\frac {-1} {(x+1)^2}} (ddxuv(\because \frac {d} {dx} \frac u v =(v.uu.v)v2)=\frac {(v.u'-u.v')} {v^2} ) So (ddx1x+1=1(x+1)2)\frac {d} {dx} \frac {1} {x+1} =\frac {-1} {(x+1)^2})

( ddxex=ex)\frac {d} {dx} e^x=e^x)



=(1)lim(x)e1(x+1)=(-1)lim(x\to\infty)e^\frac {1} {(x+1)}


=(1)e1=(-1) e^\frac {1} {\infty}


=(1)e0=(-1)e^0


=(1).1=1=(-1).1 =-1 (Ans)








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