Lim (10x - cosx + 1)/(sinx + x)
(x->0)
Lim (10x - cosx + 1)/(sinx + x)
= Lim (10 + sinx)/(cosx + 1)
= 10/2
= 5
how did you do this?
(10x - cosx + 1)/(sinx + x)
= (10 + sinx)/(cosx + 1)
remember L'Hospital's Rule:
lim f(x)/g(x) = lim f'(x)/g'(x)
when you are dealing with 0/0 or ∞/∞
Thanks :)))
To find the limit of the given expression as x approaches 0, we can use the concept of L'Hôpital's rule. L'Hôpital's 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 separately and then evaluate the limit again.
In this case, we have an indeterminate form of 0/0. Therefore, we will differentiate the numerator and denominator with respect to x.
Taking the derivative of the numerator:
lim (x->0) [10 - (-sinx)]
Since the derivative of cosx is -sinx, the derivative of -cosx is -(-sinx) = sinx.
Taking the derivative of the denominator:
lim (x->0) [cosx + 1]
The derivative of sinx is cosx.
Now we can rewrite the given expression with the derivatives:
lim (x->0) [10 - sinx] / [cosx + 1]
Now we can evaluate the limit directly by substituting x = 0 into the expression:
lim (x->0) [10 - sinx] / [cosx + 1] = [10 - sin(0)] / [cos(0) + 1]
= 10 / (1 + 1)
= 10/2
= 5
Therefore, the limit of the given expression as x approaches 0 is 5.