Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→0 (sin^−1 x)/7x
Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→0 x/(tan^−1(4x))
Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→0 (e^x−e^−x−2x)/(x−sin x)
To find the limits of the given functions, let's consider each problem separately using the given information:
1. lim x→0 (sin^−1 x)/7x
To find this limit, we can start by applying the elementary approach. Recall the basic limit: lim x→0 (sin x)/x = 1. We can use this idea in our problem.
Let u = sin^−1 x. As x approaches 0, u will also approach 0. Now, let's rewrite the limit using u.
lim x→0 (sin^−1 x)/7x = lim u→0 (u)/(7sin u)
Now we can use the basic limit mentioned earlier: lim x→0 (sin x)/x = 1. By dividing both numerator and denominator by u, we can write:
lim u→0 (u)/(7sin u) = lim u→0 (u/u)/(7sin u/u) = 1/(7lim u→0 (sin u/u))
Now, as u approaches 0, sin(u)/u approaches 1. Therefore:
lim x→0 (sin^−1 x)/7x = 1/(7*1) = 1/7
So, the limit is 1/7.
2. lim x→0 x/(tan^−1(4x))
To find this limit, we can use l'Hôpital's Rule because it involves an indeterminate form of the type 0/0.
Applying l'Hôpital's Rule, we differentiate the numerator and denominator:
lim x→0 x/(tan^−1(4x)) = lim x→0 (1)/(d/dx(tan^−1(4x)))
Differentiating the denominator with respect to x:
= lim x→0 (1)/(d/dx(tan^−1(4x)))
= lim x→0 (1)/(d/dx(arctan(4x)))
Now, we can differentiate arctan(4x) using the chain rule:
= lim x→0 (1)/(1/(1 + (4x)^2) * 4)
= lim x→0 (4)/(1 + (4x)^2)
Plugging in x = 0 into the above expression, we get:
= 4/(1 + (0)^2)
= 4/1
= 4
Therefore, the limit is 4.
3. lim x→0 (e^x−e^−x−2x)/(x−sin x)
Again, we can use l'Hôpital's Rule because it involves an indeterminate form of the type 0/0.
Applying l'Hôpital's Rule, we differentiate the numerator and denominator:
lim x→0 (e^x−e^−x−2x)/(x−sin x) = lim x→0 (d/dx(e^x−e^−x−2x))/(d/dx(x−sin x))
Differentiating the numerator and denominator with respect to x:
= lim x→0 (e^x + e^−x - 2)/(1 - cos x)
Now, we can plug in x = 0 into the above expression to find the limit:
= (e^0 + e^0 - 2)/(1 - cos 0)
= (1 + 1 - 2)/(1 - 1)
= 0/0
The numerator and denominator both evaluate to 0, which is an indeterminate form. We can continue applying l'Hôpital's Rule until we obtain a determinate form.
Second time applying l'Hôpital's Rule:
lim x→0 (d/dx(e^x + e^−x - 2))/(d/dx(1 - cos x))
= lim x→0 (e^x - e^−x)/(sin x)
Now, we can plug in x = 0 into the above expression to find the limit:
= (e^0 - e^0)/(sin 0)
= (1 - 1)/(0)
= 0/0 (still an indeterminate form)
Since we haven't obtained a determinate form yet, we need to continue differentiating. However, since we have already applied l'Hôpital's Rule twice, we can't apply it again. Therefore, we should explore other methods to solve this limit, as suggested in the problem statement.
By analyzing the limit, notice that the numerator factorizes into (e^x - 1)(1 - e^−x). Additionally, we know that lim x→0 (sin x)/x = 1.
Using these facts, we can rewrite the limit as:
lim x→0 (e^x−e^−x−2x)/(x−sin x)
= lim x→0 ((e^x - 1)(1- e^−x))/(x(1 - sin x/x))
Now, as x approaches 0, we can observe:
lim x→0 sin x/x = 1
lim x→0 e^x - 1 = 0
lim x→0 1 - e^−x = 0
Hence, we can rewrite the limit further:
lim x→0 ((e^x - 1)(1- e^−x))/(x(1 - sin x/x))
= (0)(0)/(0(1 - 1)) (by substituting the limits)
= 0/0 (still an indeterminate form)
Even after using the alternative method, we still obtain an indeterminate form. This indicates that the limit requires further analysis or a different approach.