What is the limit of [1/x] as x approaches 0?
In no case does the limit technically exist, because limits are numbers. Limits don’t give you the right language to talk about this situation accurately, but:[math]\displaystyle\frac1{0^+}\to+\infty[/math][math]\displaystyle\frac1{0^-}\to-\infty[/math][math]\displaystyle\frac1{0^{\pm}}\to\pm\infty[/math][math]\displaystyle\frac1{0}[/math] is simply undefined.These are shown in the video below.
What is the limit as x approaches 1 of x/lnx?
Denominator is zero, so the limit is infinity.
How do you evaluate [math]\lim\limits_{x\to \infty}\frac{1}{x^\frac{1}{x}}[/math]?
Let us see what happens to x^1/x as x approaches infinity. Let's define the limit as some Y.So,ln(Y)= ln(x)/xAs x approaches infinity, both x and ln(x) go to infinity. So, L'hospital's rule can be applied.So, ln(Y)=(1/x)/1As x approaches infinity, ln(Y) goes to zero.So, Y=e^0=1The required limit is 1/Y=1So, as x approaches infinity,1/(x^1/x) approaches 1
What is the limit of 1+cos(1÷x) when x approaches 0?
There is no limit. As x approaches 0 the value of [math]\cos(\frac{1}{x})[/math] oscillates between [-1,1] more and more quickly. This does not allow for the basic condition for the existence of a limit namely that there will be a segment as small as it may be arround the point (0) in this case, at which the function is bound and if we take a smaller segment around the limit point the bounds of the function are smaller in absolute value than the bounds of the larger segment. Since the value of the sine oscillates between -1,1 for any segment size around 0 no matter how small it may be this condition is not satisfied and therefore the limit does not exist.
Evaluate lim x -> 1, ((x^1/3) - 1) / ((x^1/2) - 1)?
It's probably easiest to do with l'Hopital's rule, if you know it. Since it's a 0/0 limit, just differentiate top and bottom to get an equal limit: lim x->1 [(1/3)x^(-2/3)] / [(1/2)x^(-1/2)] lim x->1 (2/3) * x^(-2/3) / x^(-1/2) lim x->1 (2/3) * x^(-1/6) = 2/3 EDIT: If you don't know l'Hopital's rule, but you do know derivatives, we can use a similar approach. We know that the derivative of x^(1/3), at x = 1, is given by the limit: lim x->1 [x^(1/3) - 1^(1/3)] / [x - 1] = lim x->1 [x^(1/3) - 1] / [x - 1] We can use power rule to compute the derivative and hence this limit. The derivative is (1/3)x^(-2/3), and so at 1, it is -2/3. Hence that limit is 1/3. Similarly, we can compute the limit: lim x->1 [x^(1/2) - 1] / [x - 1] using the derivative of x^(1/2) at x = 1 (it's 1/2). So: lim x->1 ([x^(1/3) - 1] / [x^(1/2) - 1]) = lim x->1 ([x^(1/3) - 1] / [x - 1] divided by [x^(1/2) - 1] / [x - 1]) = 1/3 divided by 1/2 = 2/3 as before. Finally, if you don't know derivatives at all, we can still work with this expression. Consider the difference of two squares and difference of two cubes formulae: u^2 - v^2 = (u - v)(u + v) u^3 - v^3 = (u - v)(u^2 + uv + v^2) If we let u = x^(1/2) and v = 1 in the difference of two squares, and u = x^(1/3) and v = 1 in the difference of two cubes, then we notice: (x^(1/2))^2 - 1^3 = (x^(1/2) - 1)(x^(1/2) + 1) (x^(1/3))^3 - 1^3 = (x^(1/3) - 1)(x^(1/3)^2 + x^(1/3) * 1 + 1^2) x - 1 = (x^(1/2) - 1)(x^(1/2) + 1) ... (1) x - 1 = (x^(1/3) - 1)(x^(2/3) + x^(1/3) + 1) ... (2) Divide (2) by (1) to obtain: (x - 1) / (x - 1) = [(x^(1/3) - 1) / (x^(1/2) - 1)] * [(x^(2/3) + x^(1/3) + 1) / (x^(1/2) + 1)] 1 = [(x^(1/3) - 1) / (x^(1/2) - 1)] * [(x^(2/3) + x^(1/3) + 1) / (x^(1/2) + 1)] (x^(1/3) - 1) / (x^(1/2) - 1) = (x^(1/2) + 1) / (x^(2/3) + x^(1/3) + 1) Therefore, as x approaches 1, the right hand side approaches (1 + 1) / (1 + 1 + 1) = 2/3, like it did with all the other methods.
Prove the limit: (ln(1 - x) - sinx)/[1 - cos^2(x)]?
======================= The Maclaurin Series of ln(1 - x) = - x - x²/2 - x³/3 - .... And the Maclaurin Series of sinx = x - x³/6 + x⁵/120 - ... When x is very small, ln(1 - x) ≈ -x, sinx ≈ x So lim(x→0) ( (ln(1 - x) - sinx )/(1 - cos²x) ) lim(x→0) ( (ln(1 - x) - sinx )/sin²x ) = lim(x→0) ( (-x) - x)/x² = lim(x→0) -2x/x² = lim(x→0) -2/x Which does not exist because lim(x→0‾) -2/x ≠ lim(x→0⁺) -2/x Confirmation: http://www.wolframalpha.com/input/?i=Mac... http://www.wolframalpha.com/input/?i=Mac... http://www.wolframalpha.com/input/?i=lim... http://www.wolframalpha.com/input/?i=limit+as+x+approaches+0+%28+ln%281+-+x%29+-+sinx%29+%2F%281+-+%28cosx%29%5E2%29
Limit x→0 (lnx)/(cotx) using l'hospital's rule?
lim (x->0+) ln x / cot x is an ∞/∞ form (note we must have x > 0 for ln x to make sense, so this is a one-sided limit). By L'Hopital's rule, if the following limit is defined it is the same as the original: = lim (x->0+) (1/x) / (-cosec^2 x) = lim (x->0+) (-sin^2 x / x) which is a 0/0 form, use l'Hopital again: = lim (x->0+) (-2 sin x cos x / 1) = 0.
Limit x tends to 0 (tanx/x)^1/x?
By L'Hopital's Rule, note that lim(x-->0) tan x/x = lim(x-->0) sec^2(x)/1 = 1. So, lim(x-->0) (tan x/x)^(1/x) is of the form "1^∞". We evaluate this using logarithms. Let L = lim(x-->0) (tan x/x)^(1/x). So, ln L = lim(x-->0) ln(tan x/x) / x, now of the form "0/0". Applying L'Hopital's Rule, we obtain ln L = lim(x-->0) [sec^2(x)/tan x - 1/x] / 1 since (d/dx) ln(tan x/x) = (d/dx) [ln(tan x) - ln x] = sec^2(x)/tan x - 1/x. Rewriting this (as this is of the form "∞ - ∞"): ln L = lim(x-->0) sec^2(x)/tan x - 1/x = lim(x-->0) 1/(sin x cos x) - 1/x = lim(x-->0) 2/sin(2x) - 1/x, still of the form "∞ - ∞" = lim(x-->0) [2x - sin(2x)] / (x sin(2x)), now of the form "0/0" Apply L'Hopital's Rule again (two times): ln L = lim(x-->0) [2 - 2 cos(2x)] / (sin(2x) + 2x cos(2x)), still of the form "0/0" = lim(x-->0) 4 sin(2x) / (4 cos(2x) - 4x sin(2x)) = 0/(4 - 0) = 0. Hence, L = e^0 = 1. I hope this helps!
What is limit x->0 sinx/x?
Let me tell you a generic result (that holds true for every value)[math]\cos^2(x) \le \dfrac{\sin(x)}x \le 1[/math]Now you might be wondering, where did it come from?So let's prove this_Now tell me what is the area of this figure? (keep in mind that the angle is in radians)If the results trouble you (you are most probably uncomfortable with radians, Google it)Let's move onAndNowDid you see that we have proved something terrible[math]\cos^2(x) \le \dfrac{\sin(x)}x \le 1[/math]Now when you are familiar with this result there is nothing left to reach your conclusion, just take the limit[math]\therefore\, \displaystyle\lim_{x\to 0} \dfrac{\sin{x}}{x}=1\tag*{}[/math]Credit: Wolfram Alfa[1]Edit:Here is the graph of [math]\dfrac{\sin{x}}{x}[/math]You can see that at [math]x=0[/math] the value is [math]1[/math]Footnotes[1] Wolfram|Alpha: Making the world’s knowledge computable