5) Use the L’Hopital’s method to evaluate the following limits. In each case, indicate what type of limit it is ( 0/0, ∞/∞, or 0∙∞)
lim x→2 sin(x^2−4)/(x−2) =
lim x→+∞ ln(x−3)/(x−5) =
lim x→pi/4 (x−pi/4)tan(2x) =
sin(x^2-4)/(x-2) --> 2xcos(x^2-4)/1 --> 4
ln(x-3)/(x-5) --> (1/(x-3))/5 -> 0
(x-pi/4)tan(2x) = (x-pi/4)/cot(2x) --> 1/(-2csc^2(2x)) = -1/2 sin^2(2x) --> -1/2
To evaluate limits using L'Hopital's method, we need to apply the rule when we have an indeterminate form of 0/0 or ∞/∞. Here's how we can use L'Hopital's method to evaluate each of the given limits:
1) lim x→2 sin(x^2−4)/(x−2):
We observe that this limit is of the form 0/0 when we substitute x=2. So, we can apply L'Hopital's rule.
Derive the numerator and denominator separately:
(sin(x^2-4))' = cos(x^2-4) ⋅ (x^2-4)' = cos(x^2-4) ⋅ 2x
(x-2)' = 1
Now, take the limit again with the derived functions:
lim x→2 (cos(x^2-4) ⋅ 2x)/(1) = cos(0) ⋅ 4 = 4
Therefore, the limit of the given function as x approaches 2 is 4.
2) lim x→+∞ ln(x−3)/(x−5):
Here, we have the form ∞/∞ when we substitute x=+∞. So, we can utilize L'Hopital's rule.
Take the derivative of the numerator and denominator:
(ln(x-3))' = 1/(x-3) ⋅ (x-3)' = 1/(x-3)
(x-5)' = 1
Now, take the limit with the derived functions:
lim x→+∞ (1/(x-3))/(1) = lim x→+∞ 1/(x-3) = 0
Thus, the limit of the given function as x approaches positive infinity is 0.
3) lim x→pi/4 (x-pi/4)tan(2x):
This limit does not result in an indeterminate form, so we don't require L'Hopital's rule. We can directly substitute x=pi/4 into the function:
(x-pi/4)tan(2x) = (pi/4-pi/4)tan(2(pi/4)) = 0⋅tan(pi/2) = 0⋅∞ = 0
Therefore, the limit of the given function as x approaches pi/4 is 0.
In summary:
- lim x→2 sin(x^2−4)/(x−2) = 4
- lim x→+∞ ln(x−3)/(x−5) = 0
- lim x→pi/4 (x−pi/4)tan(2x) = 0