it is a indeterminant form
$\lim_{X\rightarrow \infty } -(x+1)(e^{\frac{1}{x+1}}-1)$
converting into $\frac{0}{0}$ indeterminant form
$\lim_{X\rightarrow \infty } -\frac{(e^{\frac{1}{x+1}}-1)}{\frac{1}{(x+1)}}$
now this is in indeterminant form $\frac{0}{0}$ form so Lhosptal rule
differentiating num and denominator
$\lim_{X\rightarrow \infty } -\frac{(e^{\frac{1}{x+1}}*-1*\frac{1}{(x+1)^2})}{-1*\frac{1}{(x+1)^2}}$
so it will be
-$\lim_{X\rightarrow \infty } e^{\frac{1}{x+1}}$ by substituting we will get ans as -1