Evaluate these limits.
(Enter answer as a decimal. Enter INF if the limit is +∞; -INF if the limit is −∞; and DNE if the limit is neither ±∞ and does not exist.)
(a) limx→04+x−−−−−√−2x= - sin responder
(b) limx→0tan(2x)sec(3x)x
To evaluate limits, we need to substitute the given value (or approach it) into the function and see what happens. Let's evaluate the limits step by step:
(a) limx→0 4+x−−−−−√−2x
First, we substitute the limit value into the function, which gives us:
4 + 0 - √(2 * 0) = 4
Hence, the limit is 4.
(b) limx→0 tan(2x)sec(3x)/x
First, we substitute the limit value into the function:
tan(2*0)sec(3*0)/0
Simplifying this further, we get:
tan(0) * sec(0) / 0
Since tan(0) = 0 and sec(0) = 1, we have:
0 * 1 / 0 = 0 / 0
The expression 0/0 is an indeterminate form, which means we cannot determine the limit by direct substitution or simplification. In such cases, we usually need to use additional techniques such as L'Hôpital's Rule to find the limit.
To use L'Hôpital's Rule, we differentiate the numerator and denominator separately.
The derivative of tan(2x) is sec^2(2x), and the derivative of sec(3x) is 3sec(3x)tan(3x).
Now, let's re-evaluate the limit:
limx→0 [sec^2(2x) * 3sec(3x)tan(3x)] / 1
Plugging in x = 0, we get:
[sec^2(0) * 3sec(0)tan(0)] / 1
Since sec^2(0) = 1 and sec(0) = 1, the expression becomes:
[1 * 3 * 0] / 1 = 0 / 1 = 0
Therefore, the limit is 0.