Question

How do you find the limit of $x\left( {{e}^{-x}} \right)$ as x approaches infinity using L’Hospital rule?

Answer 1

Now we know that according to L’Hospital rule we have, $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$ . Now using this we will first simplify the limit and then remove the indefinite form. Now we will solve the limit substituting the value of x and hence find the solution.

Complete step by step solution:
Now consider the given function $x\left( {{e}^{-x}} \right)$ .
We want to find the limit of the function as x tends to infinity. Hence we want to find the value of $\underset{x\to \infty }{\mathop{\lim }}\,x\left( {{e}^{-x}} \right)$ .
We can also write the function as $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{x}{{{e}^{x}}}$
Let us first understand the meaning of limit. Limit of a function means the value the function will approach as x tends to a particular value. Now limits can be very useful while working on functions which are not defined over the whole domain.
Consider the example of function $f\left( x \right)=\dfrac{\left( {{x}^{2}}-4 \right)}{x-2}$
Now we know that the function is not defined at x = 2.
But let us check the value of limit $\underset{x\to 2}{\mathop{\lim }}\,\dfrac{{{x}^{2}}-4}{x-2}=\underset{x\to 2}{\mathop{\lim }}\,\dfrac{\left( x-2 \right)\left( x+2 \right)}{x-2}=\underset{x\to 2}{\mathop{\lim }}\,\left( x+2 \right)=4$
Hence even though the function does not exist at x = 2 we can say that just before and after x = 2 the value of function is 4.
Now to find the limits of the given function we will use the L'Hospital rule.
Now L’Hospital rule states that the limits of a function $\dfrac{f\left( x \right)}{g\left( x \right)}$ is same limits of the function $\dfrac{f'\left( x \right)}{g'\left( x \right)}$ .
Now consider the given function $\dfrac{x}{{{e}^{x}}}$ . comparing the function with $\dfrac{f\left( x \right)}{g\left( x \right)}$ we get, $f\left( x \right)=x$ and $g\left( x \right)={{e}^{x}}$ . Now differentiating the functions we get, $f'\left( x \right)=1$ and $g'\left( x \right)={{e}^{x}}$ .
Now we know that $\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{f\left( x \right)}{g\left( x \right)}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{f'\left( x \right)}{g'\left( x \right)}$ .
Hence we have,
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{x}{{{e}^{x}}}=\underset{x\to \infty }{\mathop{\lim }}\,\dfrac{1}{{{e}^{x}}}$
Now we know that as $x\to \infty$ , ${{e}^{x}}\to \infty$ hence we have,
$\Rightarrow \underset{x\to \infty }{\mathop{\lim }}\,\dfrac{x}{{{e}^{x}}}=0$
Hence the value of the given limit is 0.

Note: Now note that we can apply L’Hospital rule several times till we get a form of limit which is solvable. Also note that we do not differentiate the whole fraction as a function using the division rule of differentiation but we differentiate the numerator and denominator separately.