Answer
$\frac{1}{2}$
Work Step by Step
$\lim\limits_{l \to +\infty}$$(-\frac{1}{2}e^{-2x})$ for x from 0 to l
To calculate this limit, we first put the value of x from 0 to l to calculate the value of the expression within the bracket. The $\frac{1}{2}$ will come out of the bracket because it will remain unaffected from the change in the limits. Because $\frac{1}{2}$ is just a number and is not affected by change in the value of l, we can safely take it out of the bracket. Hence the expression will change to the one given below.
= $\frac{1}{2}$$\lim\limits_{l \to \infty} (-e^{-2l} + 1)$
Now, the expression becomes $(-e^{-2l} + 1)$. We have to calculate what this expression would converge to when the value of $l$ tends to +$\infty$
Again here, 1 will remain the same. It is not affected even when l becomes 0 or $\infty$; thus, we can focus our attention to the first term.
We know that for $e^{x}$
When
x=0 ; $e^{x}$=1
x= +$\infty$ $e^{x}$= $\infty$
x = -$\infty$ $e^{x}$ = 0
We know that for $e^{-x}$
When
x=0 ; $e^{-x}$=1
x= +$\infty$ $e^{-x}$= 0
x = -$\infty$ $e^{-x}$ = $\infty$
Considering the above,
The expression is reduced to $\frac{1}{2}$(0+1) = $\frac{1}{2}$
