Evaluate the limits lim x^3 + 3/ x->¤ 2x^3 + 4x + 1 lim 1 - cosx / x->o x^2
Use l'Hôpitals rule:
1.
Lim (x->∞) (x³+3)/(2x³+4x+1)
= Lim (x->∞) (3x²)/(6x²)
= Lim (x->∞) (6x)/(12x)
= Lim (x->∞) 1/2
= 1/2
Similarly for (1-cos(x))/(x²)
= Lim (x->∞)
To evaluate these limits, let's take them one by one.
1. lim (x^3 + 3) / (2x^3 + 4x + 1) as x approaches infinity (represented as "¤").
To find the limit as x approaches infinity, we consider the highest power of x in the numerator and denominator, which is x^3 for both. By dividing each term in the expression by x^3, we get:
lim (x^3 + 3) / (2x^3 + 4x + 1) = lim (1 + 3/x^3) / (2 + 4/x^2 + 1/x^3) as x approaches infinity.
As x approaches infinity, the terms with 1/x^3 and 1/x^2 become negligible compared to the terms with x^3 and x^2. Therefore, we can simplify the expression:
lim (x^3 + 3) / (2x^3 + 4x + 1) ≈ lim (1 + 0) / (2 + 0 + 0) = 1/2.
So, the limit of the given function as x approaches infinity is 1/2.
2. lim (1 - cos x) / x^2 as x approaches zero (represented as "o").
To evaluate this limit, we can apply L'Hôpital's rule. Take the derivative of both the numerator and denominator with respect to x:
lim (1 - cos x) / x^2 = lim (-sin x) / (2x) as x approaches zero.
Now, we can substitute x = 0 into the resulting expression:
lim (-sin x) / (2x) = lim (-sin 0) / (2 * 0) = 0 / 0.
We obtained the indeterminate form "0 / 0," which means we need to apply L'Hôpital's rule again. Differentiate the numerator and denominator:
lim (-sin x) / (2x) = lim (-cos x) / 2 as x approaches zero.
Now we can substitute x = 0 into the new expression:
lim (-cos x) / 2 = (-cos 0) / 2 = -1 / 2.
Therefore, the limit of (1 - cos x) / x^2 as x approaches zero is -1/2.