Help solving a limit?????
multiply and divide with √8(a+h) + √8a so lim.........sqrt(8(a+h)) - sqrt(8a) h->0......----------------------------... ..............................h =lim h--->0 (√8(a-h) -√8a)(√8(a+h) + √8a)/h(√8(a+h) + √8a) =lim h--->0 [√8(a-h)]² -[√8a]²/h(√8(a+h) + √8a) =lim h--->0 [8a-8h -8a]/h(√8(a+h) + √8a) =lim h--->0 -8h/h(√8(a+h) + √8a) =lim h--->0 -8/(√8(a+h) + √8a) =-8/(√8(a+0) + √8a) =-8/2√8a = -4/√8a = -4/ 2√2a = -2/√2a
Help solving a limit...?
[x→0]lim x csc (2x)/cos (5x) This is NOT 0/0 form, so just use the main limit theorem: ([x→0]lim x csc (2x)) / ([x→0]lim cos (5x)) ([x→0]lim x csc (2x)) / 1 [x→0]lim x csc (2x) [x→0]lim x/sin (2x) This IS a 0/0 form, so use L'hopital: [x→0]lim 1/(2 cos (2x)) 1/2 Done.
I need help solving this limit Lim x->0- sqrt[x^4 * cos^2(3x) + x^2 * sin^2(x/2) - x^4]/[x^2]?
Using power series for the trig. functions (using half angle identities first): lim(x→0-) √[x⁴ cos²(3x) + x² sin²(x/2) - x⁴] / x² = lim(x→0-) √[x⁴ * (1/2)(1 + cos(6x)) + x² * (1/2)(1 - cos x) - x⁴] / x² = lim(x→0-) √[(1/2) (x⁴ (1 + cos(6x)) + x² (1 - cos x) - 2x⁴)] / x² = lim(x→0-) √(1/2) * √[x⁴ (1 + cos(6x)) + x² (1 - cos x) - 2x⁴] / x² = (1/√2) * lim(x→0-) √[x⁴(1 + (1 - (6x)²/2! + ...)) + x²(1 - (1 - x²/2! + ...)) - 2x⁴]/x² = (1/√2) * lim(x→0-) √[x⁴(2 - 18x² + ...) + (x⁴/2 - ...) - 2x⁴]/x² = (1/√2) * lim(x→0-) √[x⁴ ((2 - 18x² + ...) + (1/2 - ...) - 2)]/x² = (1/√2) * lim(x→0-) x² √(1/2 + (x-terms)) / x² = (1/√2) * lim(x→0-) √(1/2 + (x-terms)) = (1/√2) * √(1/2 + 0) = 1/2. Double check: http://www.wolframalpha.com/input/?i=lim... --------- Update: Without using series or L'Hopital's Rule. lim(x→0-) √[x⁴ cos²(3x) + x² sin²(x/2) - x⁴] / x² = lim(x→0-) √[x⁴ (cos²(3x) + sin²(x/2)/x² - 1)] / x², by factoring = lim(x→0-) x² √[cos²(3x) + sin²(x/2)/x² - 1] / x², = lim(x→0-) √[cos²(3x) + sin²(x/2)/(2x/2)² - 1] = lim(x→0-) √[cos²(3x) + (1/4) (sin(x/2)/(x/2))² - 1] = √[1² + (1/4) 1² - 1], via lim(t→0) sin(t)/t = 1 with t = x/2 = 1/2. ------- I hope this helps!
Consider the following limit (need help solving)?
When x=-2, the denominator is 0, so the only way the fraction could reach a limit is if the numerator also equaled 0 at x=-2 (thus producing and indeterminate form). So, 4x^2 + ax + a + 4 must be 0 at x=-2: 4*4 - 2a + a + 4 = 0, 16 + a + 4 = 0, a = -20 So a = -20 for the limit to exist. To find the value of the limit, see that the numerator's dominant term is 4 times that of the denominator, so the limit will be equal to 4. So... lim (4x^2 + ax + a + 4)/(x^2 + x - 2) = 4, if a = -20 x->-2
Help solving a limit as n approaches infinity?
n^p is a polynomial and e^n is an exponential. An exponential grows much much faster then a polynomial. Hence, as n approaches infinity, e^n will approach infinity faster than n^p. If the denominator goes to infinity first then (n^p/e^n) would approach zero. I think this is also one of the standard limits. If the question had been e^n/n^p then the limit have would been undefined.
I need the solution to this limit. How can I solve it?
You are supposed to apply the L'Hopital rule when you get a 0/0 form. So you are not actually stuck. The question itself is telling you what to do. Now try the question before reading rest of the answer.Now lets try to solve this problem. I rewrite tan in terms of cos for simplicity and we get[math]\frac{1-x}{1-\cos(x)}-\frac{(2+x)\cos^2(x)}{1-\cos^2(x)}[/math][math]=\frac{(1-x)(1+\cos(x))-(2+x)\cos^2(x)}{1-\cos^2(x)}[/math]This term is in 0/0 form when we put [math]x=0[/math]. Hence we differentiate the numerator and the denominator and get:[math]\frac{-(1+\cos(x))+(1-x)(-\sin(x))-\cos^2(x)+(2+x)\sin(2x)}{\sin(2x)}[/math]Here the numerator has a value of -3 whereas the denominator has a value of 0. So we conclude that the limit doesn't exist. The expression goes towards infinity.
Need help solving a the limit of an exponential function?
The velocity v(t) of a 179 pound skydiver is approximated by v(t)= 179/1.1 * ( 1 − e^(−24.2 t/179) ) What is the skydiver's terminal velocity? That is, in terms of the constants of the problem the limit as t approaches infinity. Also, how long after jumping does it take the skydiver to reach 89 percent of the terminal velocity? I am so stuck so any help would be appreciated.
How do you solve this limit problem ? Please Help !! "Step By Step "?
sin x + cos x = sqrt(2) ( sin x/sqrt(2) - cos x /sqrt(2)) = sqrt(2) ( sin x cos pi/4 - cos x sin pi/4 = sqrt(2) ( sin ( x - pi/4) so ( sin x - cos x ) / ( pi - 4x) = sqrt(2) sin ( x - pi/4) / ( -4 ( x - pi /4 ) = - sqrt(2) / 4 * sin (x-pi/4)/(x-pi/4) = - sqrt(2)/4 *1 as x -> pi/4 or x - pi/4 -> 0 = - sqrt(2) / 4
Why do we need limits in mathematics?
I assume the question asker is referring to limit of functions and not the philosophical limitation of mathematics. I like the answer Sawarnik gave as it highlights the importance in calculus and I'd like to expand his answer a bit and hopefully provide some historical motivation.The need for limits isn't always apparent but I'll give you a classical example from the Greeks. Zeno was a philosopher who came to Plato with a set of paradoxes (many of which involved infinitesimals). The essence of the Zeno paradoxes is "that which is in locomotion must arrive at the half-way stage before it arrives at the goal." Since we can keep subdividing the distance an infinite number of times, one would never reach the goal.Of course, this isn't true and Aristotle and Plato and a host of others had their hand at resolutions to the Zeno paradox, but its really calculus and the introduction of limits that gives the simple pre-calculus tools to handle such a situation. It turns out that with a bit of calculus you can show that the series[math] \frac12+\frac14+\frac18+...=\sum_{i=1}^N \frac{1}{2^i} [/math]converges to one. We have many many more examples of such situations where limits play an important role in calculus. Calculus, in many ways, can be thought as infinitesimal analysis that easily handles the problems Zeno raises. An infinitesimal is an indefinitely small quantity which should be defined formally. En route to defining the infinitesimal one, again, encounters the necessity of the limit. Its quite beautiful mathematics when you get your hands dirty with proof based calculus courses (e.g. (ε, δ)-definition of limit). Limits are also important for considering the of behavior of functions when their inputs get large as in Asymptotic analysis. This is the dominant method for analyzing the efficiency for analyzing computational efficiency or understanding how small perturbations affect the dynamics of physical systems.In the immortal words of the Mighty Mos Def: "Numbers is hard and real and they never have feelings. But you push too hard, even [sequences of] numbers got limits."