Please help me with the two given questions.
1] find lim 3x^2-4x/7
x-->0
2] lim sinx/2x
x-->0
1) 0 - 0 = 0
2) (1/2)
Thank you- drwls!!!
Sure! I can help you with both of these questions. Let's start with the first one.
1) To find the limit of (3x^2 - 4x)/7 as x approaches 0, we can plug in the value of 0 into the expression and simplify.
Substituting x = 0 into the expression gives us (3(0)^2 - 4(0))/7 which simplifies to 0/7.
To determine the limit, we need to consider what happens to the expression as x gets very close to 0. In this case, the expression simplifies to 0/7, which equals 0.
Therefore, the limit of (3x^2 - 4x)/7 as x approaches 0 is 0.
Now let's move on to the second question.
2) To find the limit of sin(x)/(2x) as x approaches 0, we can again substitute the value of 0 into the expression and simplify.
Substituting x = 0 into the expression gives us sin(0)/(2(0)), which simplifies to 0/0.
However, we cannot directly evaluate the limit with this indeterminate form. To solve this, we can use L'Hopital's rule. This rule states that if we have an indeterminate form of 0/0 or ∞/∞, we can take the derivative of the numerator and denominator and then evaluate the limit again.
Taking the derivative of sin(x) with respect to x gives us cos(x), and differentiating 2x with respect to x gives us 2.
Now we can evaluate the limit again using the new derivatives: lim x-->0 of cos(x)/2.
Substituting x = 0 into the new expression gives us cos(0)/2, which simplifies to 1/2.
Therefore, the limit of sin(x)/(2x) as x approaches 0 is 1/2.
In summary, the limit of (3x^2 - 4x)/7 as x approaches 0 is 0, and the limit of sin(x)/(2x) as x approaches 0 is 1/2.