Limit as x approaches zero (1/x-1/tanx)
massage the form to
1/x - cosx/sinx
= (cosx-x)/(x*sinx)
and apply l'Hospital's Rule
That is where I am confused ...
Oops. A typo. Should be
(x*cosx-sinx)/(x*sinx)
just take derivatives, top and bottom. The limit is the same as for
(-x*sinx+cosx-cosx)/(sinx + xcosx)
= (-x sinx)/(sinx + x cosx)
still -> 0/0, so do it again:
(-sinx - x*cosx)/(cosx + cosx - x*sinx)
-> 0/2 = 0
To find the limit as x approaches zero of the expression (1/x - 1/tan(x)), we can analyze the two terms separately and then combine the results.
Let's start by analyzing the first term, 1/x. As x approaches zero, the value of 1/x will tend to positive or negative infinity. This is because when we divide 1 by numbers that are getting closer and closer to zero, the result becomes larger and larger in magnitude.
Next, let's consider the second term, 1/tan(x). As x approaches zero, the value of tan(x) also approaches zero. Recall that the tangent function has a vertical asymptote at x = π/2 and x = -π/2, where it approaches positive or negative infinity. This means that as x gets very close to zero, tan(x) becomes very large in magnitude, either positive or negative infinity.
Now, let's combine the two terms, (1/x - 1/tan(x)). As both terms tend towards infinity, we can rewrite the expression as (infinity - infinity). In this case, the subtracted infinities cancel each other out, and the result may be indeterminate.
To determine the actual limit, we can use algebraic manipulation. Multiply both terms by the conjugate of their denominators to eliminate the fractions. The expression can be rewritten as:
[(tan(x) - x) / (x * tan(x))]
Now, as x approaches zero, we end up with (0/0) which is an indeterminate form. This is where further analysis is needed.
To proceed, we can apply L'Hospital's Rule. This rule states that if we have an indeterminate form of the type (0/0) or (∞/∞), we can take the derivative of the numerator and denominator independently until we obtain a determinate form. Let's differentiate the expression:
[1 - 1 / (sec^2(x) * x) ] / (tan(x) + x*sec^2(x))
Here, we have used the derivative of tan(x) = sec^2(x) and the derivative of 1/x = -1/x^2.
Now, we substitute x=0 into the expression and simplify:
[1 - 1/ (sec^2(0) * 0)] / (tan(0) + 0 * sec^2(0))
[1 - 1/ (1 * 0)] / (0 + 0 * 1)
[1 - 1/0] / (0)
This is still an indeterminate form of the type "∞ - ∞".
To further analyze this expression, we need to simplify it even more. We can use the factorization technique to rewrite the numerator:
[(sec^2(x) - x) / (x * sec^2(x))]
Now, we substitute x=0 into the simplified expression:
[(1 - 0) / (0 * 1)]
[1 / (0)]
This is an indeterminate form of the type "1/0", which tends to infinity. Therefore, the limit as x approaches zero of (1/x - 1/tan(x)) is positive infinity.