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Calculus Evaluate The Limits

Evaluate the limit? CALCULUS?

lim [x→∞] x tan⁻¹(x/2) = ∞ * π/2 = ∞


lim [x→∞] x tan⁻¹(2/x) = ∞ * 0 ----> indeterminate

To solve rewrite as tan⁻¹(2/x) / (1/x), which will give 0/0
This will allow us to use L'Hopital's Rule:

lim [x→∞] x tan⁻¹(2/x)
= lim [x→∞] tan⁻¹(2/x) / (1/x)
= lim [x→∞] (−2/x² * 1/(1+(2/x)²)) / (−1/x²)
= lim [x→∞] 2/(1+(2/x)²))
= 2/(1+0)
= 2

Evaluate the limit. (csc(x) - cot(x)) [Calculus]?

csc x - cot x = (1/sin x) - (cos x/sin x) = (1 - cos x) / sin x

Use L'Hopital's:

lim(x->0) (1-cos x) / sinx
= lim(x->0) (sin x) / (cos x)
= 0

CALCULUS: EVALUATE FOLLOWING LIMITS?

for the first limit, we have form 0/0 and can use L'Hopital's Rule.
taking derivatives of top/bottom independently we get [4sin4x-3sin3x]/2x
we still have form 0/0 and do it again: [16cos4x-9cos3x]/2. Plug in 0 and get 7/2 as our limit.
Now for question 2; As x->oo, we have form n/oo where |n|<2 since cosnx varies between -1 and 1. so we see that the expression approaches 0 as x increases without bound.

Calculus Evaluate The Limits?

1. Evaluate: lim 5 - 2x^3 / x^2 + 1 x---> positive infinity and x----> negative infinity
lim(x-->+∞) (5 - 2x³) / (x² + 1) = lim(x-->+∞) [ (5/x³ - 2) / (1 + 1/x²) ] * (x³/x²) = [ (0 - 2) / (1 + 0) ] * lim(x-->+∞) x = -2 * lim(x-->+∞) x = -∞
lim(x-->-∞) (5 - 2x³) / (x² + 1) = lim(x-->-∞) [ (5/x³ - 2) / (1 + 1/x²) ] * (x³/x²) = [ (0 - 2) / (1 + 0) ] * lim(x-->-∞) x = -2 * lim(x-->-∞) x = +∞


2. Evaluate: lim (2x^3 - 100x + 5) x---> positive infinity
lim(x-->+∞) (2x³ - 100x + 5) =lim(x-->+∞) (2 - 100/x² + 5/x³) * lim(x-->+∞) x³ =(2 - 0 + 0) * lim(x-->+∞) x³ =+∞

3. Evaluate: lim (2 + 3^-x) x---> positive infinity
lim(x-->+∞) (2 + 3⁻ˣ) =lim(x-->+∞) (2 + 1/3ˣ) = 2+0=2

4. Evaluate the lim f(x) = x+2 , x not equal to 1 and 10 , x is equal to 1 x--->1
lim(x-->1) x+2 = 2 + lim(x-->1) x=3

5. Evaluate the lm (t+7)^2 - 49 / t t---> 0
lim(t-->0) ((t+7)² - 49) / t =lim(t-->0) (t²+14t+49 - 49) / t = lim(t-->0) (t+14) =14

6. Evaluate: lim cos^2 y / 2tany y --> pi/4
lim(y-->π/4) cos² y / (2tan(y)) = cos² (π/4) / (2tan(π/4)) = ¼

7. Evaluate: lim pi^e y-->2
lim(y-->2) π^e = (π^e) * lim(y-->2) 1 =π^e

I need help evaluating the limits-Calculus 2?

u must b aware of the standard limits, two of them can be used to solve these questions
Lim x→0 (1+x)^(1/x) = e
and
Lim x→∞ (1 + (1/x))^(x) = e

using these standard limist we can evaluate the above limits

1. Lim x→0 (1-3x)^(4/x)
= Lim x→0 ((1 - 3x)^(-1/3x))^(-3 x 4) ....(using the first property)
we take negtive sign cuz x has negtive sign in the base
= [Lim x→0 (1 - 3x)^(-1/3x)]^(-12)
= 1/(e^12)

2. Lim x→∞ (12x/(12x+6))^(8x)
= Lim x→∞ (1/(1 + (1/2x)))^(8x)
= Lim x→∞ (1)/((1 + (1/2x))^2x)^4 ....(using second property)
= 1/(e^4)

CALCULUS- Evaluate the following limit. Lim as x -> infinity. (e^(4x))/(x^2)?

It is infinity since the exponential function grows much faster than any polynomial function.

You can also use the L'Hospistal theorem.

BC Calculus- Evaluate a limit and derivative question?

Evaluate this limit:
lim as x-->0 of (5x)/(3sinxcosx) WITHOUT USING CALCULATOR. i used a calc and my answer is 5/3 but teacher wants us to "show work". how do u do this by hand? also...

Let g(x)= f(x^2 +1). Find g'(x). this is derivative question thanx

Calculus limit laws help?

7. Evaluate the limit using the appropriate Limit Law(s). (If it does not exist, enter NONE.)
lim_(x->-2) (4 x^2 + 4)/(x^2 + 3 x - 2)

8. Evaluate the limit using the appropriate Limit Law(s). (If it does not exist, enter NONE.)
lim_(x->8) (1 + root3(x))(4 - 2 x^2 + x^3)

9. Evaluate the limit using the appropriate Limit Law(s). (If it does not exist, enter NONE.)
lim_(t->-2) (t^2 - 3)^3(t + 3)^5

10. Evaluate the limit using the appropriate Limit Law(s). (If it does not exist, enter NONE.)
lim_(x->1) ((1 + 3 x) / (2 + 4 x^2 + 2 x^4))^3

11. Evaluate the limit, if it exists. (If it does not exist, enter NONE).
lim x→5 x2 − 5x/x2 − 4x − 5

12. Evaluate the limit, if it exists. (If it does not exist, enter NONE).
lim_(t->-8) (t^2 - 64) / (2t^2 + 17 t + 8)

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