lim((2*exp(x)+x)/(exp(x)+1), x->+inf)
1) Let's evaluate the limit of 2exp(x) + xexp(x) + 1 when x+∞
The limit of exp(x) is +∞ when x+∞
The limit of 2exp(x) is +∞ when x+∞
The limit of 2exp(x) + x is +∞ when x+∞
The limit of exp(x) + 1 is +∞ when x+∞
For 2exp(x) + xexp(x) + 1, we encounter the indeterminate form +∞+∞ as x tends to +∞
2) To find the limit, we divide the numerator and the denominator by exp(x):
Let's reduce: 2exp(x) + xexp(x)
2 + xexp(x)
3) Let's reduce: exp(x) + 1exp(x)
1 + 1exp(x)
We obtain a new fraction whose limit can be determined: 2 + xexp(x)1 + 1exp(x)
For xexp(x), we encounter the indeterminate form +∞+∞ as x tends to +∞
By comparative growth, exp(x) grows faster than x as x+∞:
The limit of xexp(x) is 0⁺ when x+∞
The limit of 2 + xexp(x) is 2 when x+∞
The limit of 1exp(x) is 0⁺ when x+∞
The limit of 1 + 1exp(x) is 1 when x+∞
The limit of 2 + xexp(x)1 + 1exp(x) is 2 when x+∞
So the limit of 2exp(x) + xexp(x) + 1 is 2 when x+∞
Answer:   2