Question

How do you know when to use L’Hospital’s rule twice?

Answer 1

In the above question, the concept is based on the concept of trigonometry. The main approach towards solving the above expression is by using trigonometric identities. The expression has cosecant function which can be converted into other trigonometric functions.

Complete step-by-step solution:
We can know when to use L’Hospital’s rule twice as follows
As soon as you try substitution in the limit and you are getting $\dfrac{0}{0}\;{\text{or}}\;\dfrac{\infty }{\infty }$, L'Hospital rule may be used there. Let us understand this with an example,
$\Rightarrow L = {\lim _{x \to 0}}\dfrac{{{e^x} - x - 1}}{{{x^2}}}$
Now substitution method directly and substitute $x = 0$ we will get
$\Rightarrow L = \dfrac{{1 - 0 - 1}}{0} = \dfrac{0}{0}$
Here we are getting $\dfrac{0}{0}$, so we can apply L'Hospital rule,
After applying we will get,
$\Rightarrow L = {\lim _{x \to 0}}\dfrac{{{e^x} - 1}}{{2x}}$
Now again try to directly substitute the value of $x = 0$ we will again get $\dfrac{0}{0}$
$\Rightarrow L = \dfrac{{1 - 1}}{{2 \times 0}} = \dfrac{0}{0}$
So applying L'Hospital rule once again, we will get
$\Rightarrow L = {\lim _{x \to 0}}\dfrac{{{e^x}}}{2}$
Now again trying substitution method that is substituting the value of $x = 0$
$\Rightarrow L = \dfrac{{{e^0}}}{2} = \dfrac{1}{2}$
So we have used the L’Hospital rule twice here in this example, hence we have to see whether the condition is satisfying to apply this rule or not then we will apply the L'Hospital rule.

Note: L’Hospital rule can be used as many times as possible depending on how long the indeterminate form $\dfrac{0}{0}\;{\text{or}}\;\dfrac{\infty }{\infty }$ breaks, also with this the above four conditions should also be satisfied in order to use L’Hospital rule one time, two times or as many times as needed.