Question

Lim_x→∞ -(x+1)(e^1/(x+1) -1)

Answer 1

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

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

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

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

$=(-1)lim(x\to\infty)\frac {e^\frac {1} {(x+1)}. \frac {-1} {(x+1)^2}} {\frac {-1} {(x+1)^2}}$ $(\because \frac {d} {dx} \frac u v$ $=\frac {(v.u'-u.v')} {v^2} )$ So ($\frac {d} {dx} \frac {1} {x+1} =\frac {-1} {(x+1)^2})$

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

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

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

$=(-1)e^0$

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