Welcome to the last key topic of unit 4! 🥳
You may recall from Unit 1 that sometimes the limits of functions evaluate to ±∞∞ or 00 which are indeterminate forms. Now, instead of looking for another way to manipulate the equation to try and find an answer that is not in indeterminate form, we can use our knowledge of derivatives to help us! Specifically, we can use L’Hopital’s Rule. ⭐
L’Hopital’s Rule states that if limx→ag(x)f(x)=00 or limx→ag(x)f(x)=±∞∞,
x→alimg(x)f(x)=x→alimg′(x)f′(x)
Basically, the rule states that we can evaluate the limits of indeterminate forms using their derivatives!
Let’s try a practice problem together! Evaluate the following limit.
x→2πlimx−2πcos(x)
Plugging x=2π into x−2πcos(x) results in the indeterminate form 00. This signals to us that we should use L’Hopital’s Rule.
When completing a free-response question on the AP exam, it is important to show the limits of f(x), the numerator, and g(x), the denominator, are separately equal to the needed parameters. Let’s go ahead and do that now!
x→2πlimcos(x)=0
x→2πlimx−2π=0
Since limx→2πcos(x)=0 and limx→2πx−2π=0, L’Hopital’s Rule can be applied. Be sure to write this statement out before actually applying this rule.
Now, we can take the derivatives and get into L’Hopital’s Rule.
x→2πlimx−2πcos(x)
x→2πlimdxd[x−2π]dxd[cos(x)]
x→2πlim1−sin(x)
=1−sin(2π)
In conclusion, we know that this limit…
x→2πlimx−2πcos(x)=−1
Great work! Doesn’t L’Hopital’s Rule save so much time? 🙃
Here are some problems you can practice applying L’Hopital’s Rule on!
Evaluate the following limits. Imagine these are free-response questions and you have to check conditions!
Question 1: L’Hopital’s Rule
x→0lim7x+tan(x)tan(x)
Question 2: L’Hopital’s Rule
x→∞lim7x2+213x2−8
Question 1: L’Hopital’s Rule
Plugging x=0 into 7x+tan(x)tan(x) results in the indeterminate form 00. Since the expression involves mixed function types, it is not possible to manipulate it algebraically in any way to find the limits. Therefore, we should use L’Hopital’s Rule.
But first, show that the limits are separately equal to 0.
x→0limtan(x)=0
x→0lim7x+tan(x)=0
Since limx→0tan(x)=0 and limx→07x+tan(x)=0, L’Hopital’s Rule can be applied.
x→0lim7x+tan(x)tan(x)
=x→0limdxd[7x+tan(x)]dxd[tan(x)]
=x→0limx+sec2(x)sec2(x)
=x+sec2(0)sec2(0)
=x+sec2(0)1
In conclusion…
x→0lim7x+tan(x)tan(x)=81
Question 2: L’Hopital’s Rule
Plugging x=∞ into 7x2+213x2−8 results in the indeterminate form ∞∞. Therefore, we should use L’Hopital’s Rule.
x→∞lim3x2−8=∞
x→∞lim7x2+21=∞
Since limx→∞3x2−8=∞ and limx→∞7x2+21=∞, L’Hopital’s Rule can be applied.
x→∞lim7x2+213x2−8
=x→∞limdxd[7x2+21]dxd[3x2−8]
=x→∞lim14x6x
Therefore…
x→∞lim7x2+213x2−8=73
Remember, L'Hopital's Rule provides a powerful method for evaluating limits involving indeterminate forms. By taking derivatives of the numerator and denominator, it helps simplify complex limits, leading to easier evaluation.
Happy calculus studying! 📚