Lim sin2h sin3h / h^2
h-->0
how would you do this ?? i got 6 as the answer, just want to make sure it's right.
and i couldn't get this one (use theorem 2)
lim tanx/x
x-->0
and also this one (use squeeze theorem to evaluate the limit)
lim (x-1)sin Pi/x-1
x-->1
the first one is correct.
Here is a simple way to check your limit answers if you have a calculator
pick a value very "close" to your approach value, in this case I would pick x = .001
evaluate using that value, (you are not yet dividing by zero, but close)
I got a value of 5.999985 which I would say is close to 6
for lim tanx / x as x -->0
= lim (sinx/cosx)/x
= lim (sinx/x)*lim 1/cosx as x ----> 0
= 1*1 = 1
If you meant lim (x-1)sinπ/(x-1)
wouldn't the last one simply be sinπ or 0 ?
To find the limit of the function lim(sin(2h)sin(3h)/h^2) as h approaches 0, you can use the limit definition of derivative or the properties of sine function.
Let's break it down step by step:
1. Rewrite the expression: lim(sin(2h)sin(3h)/h^2) = lim(sin(2h)/h) * lim(sin(3h)/h) * lim(1/h)
Notice that we have used the property sin(A) / B = sin(A/B) in the first two limits.
2. Evaluate each individual limit:
lim(sin(2h)/h) as h approaches 0:
To evaluate this limit, you can use the limit definition of derivative:
lim(sin(2h)/h) as h approaches 0 is equal to the derivative of sin(2h) with respect to h, evaluated at h=0.
Taking the derivative, we get: d/dh(sin(2h)) = 2cos(2h).
Evaluating at h=0, we have: 2cos(0) = 2.
Therefore, lim(sin(2h)/h) as h approaches 0 is equal to 2.
lim(sin(3h)/h) as h approaches 0:
Following the same steps as above, we find that this limit is also equal to 3.
lim(1/h) as h approaches 0:
This limit goes to infinity since the denominator approaches 0.
3. Combine the results:
lim(sin(2h)/h) * lim(sin(3h)/h) * lim(1/h) = 2 * 3 * infinity = infinity.
Therefore, the limit of sin(2h)sin(3h)/h^2 as h approaches 0 is infinity, not 6 as you initially suggested.
Regarding the second question, to find the limit of tan(x)/x as x approaches 0, you can use L'Hospital's Rule or a trigonometric identity.
1. Using L'Hospital's Rule:
Take the derivative of both the numerator and denominator separately, then evaluate the limit again:
lim(tan(x)/x) as x approaches 0 = lim(sec^2(x)/1) as x approaches 0 = sec^2(0) = 1.
2. Using a trigonometric property:
tan(x) can be written as sin(x)/cos(x):
lim(tan(x)/x) as x approaches 0 = lim(sin(x)/(x*cos(x))) as x approaches 0.
We can simplify this using the squeeze theorem:
-1 <= sin(x)/(x*cos(x)) <= 1 for all x close to 0.
Therefore, by the squeeze theorem, the limit is 1.
So, the limit of tan(x)/x as x approaches 0 is 1.
Regarding the third question, to evaluate the limit of (x-1)sin(Pi/(x-1)) as x approaches 1 using the squeeze theorem:
1. First, find the bounds for the function:
The sine function is bounded between -1 and 1 for any value of its argument.
So, -1 <= sin(Pi/(x-1)) <= 1.
Now multiply each side by (x-1):
-(x-1) <= (x-1)sin(Pi/(x-1)) <= (x-1).
2. Evaluate the limits where x approaches 1:
Consider each side of the inequality separately.
lim(-(x-1)) as x approaches 1:
This is simply -1.
lim((x-1)) as x approaches 1:
This is also equal to 0.
3. Apply the squeeze theorem:
Since -1 <= (x-1)sin(Pi/(x-1)) <= (x-1) for all x close to 1, and the limits of the left and right sides are -1 and 0, respectively, by the squeeze theorem, the limit of (x-1)sin(Pi/(x-1)) as x approaches 1 is equal to 0.
Therefore, the limit of (x-1)sin(Pi/(x-1)) as x approaches 1 is 0.