Tag Info

Hot answers tagged limits-without-lhopital

29

Limits aren't radio dials. $$\lim \left(1 +\color{blue}{\frac 1x}\right)^{\color{orange}x}$$ You can't tweak the $\color{blue}{\text{blue}}$ limit first to get it down to $1 + \color{blue}0$ and then tweak the $\color{orange}{\text{orange}}$ limit second to get $(1+ \color{blue}0)^{\color{orange}\infty} = 1$ Food for thought: Why don't you tweak the ...

6

$\tan x-\sin x=\sin x(1/\cos x-1)=\sin x(1-\cos x)/\cos x$ Now notice that continuing now from where you got stuck you can group terms like $$\frac{\sin(x)}x \cdot \frac{1-\cos x}{x^2} \cdot \frac1{ \cos x (\sqrt{1+\tan x}+\sqrt{1+\sin x}}$$.

5

Your limit can be found using the series expansions and hence asymptotics of each function near $x=0$. \begin{align} \lim_{x\to0^+}\left(\frac{\cos^2{(x)}}{x}-\frac{e^x}{\sin{(x)}}\right) &=\lim_{x\to0^+}\left(\frac{\sin{(x)}\cos^2{(x)}-xe^x}{x\sin{(x)}}\right)\\ &=\lim_{x\to0^+}\left(\frac{\sin{(x)}(1+\cos{(2x)})-2xe^x}{2x\sin{(x)}}\right)\\ &=... 5 One way Apply definition of derivative of 2^x\Longrightarrow \lim_{x\to 0} \frac{2^x-2^0}{x-0}=\frac{d}{dx}(2^x)|_0=2^0\log 2\Longrightarrow \lim_{x\to 0}\frac{x}{2^x-1}=\frac{1}{\log 2}$$Second way Apply change of variable, let 2^x=y$$\Longrightarrow x=\frac{\log y}{\log 2}$$Also, as x\rightarrow 0, y\rightarrow 1. So, required ... 5 You can apply the binomial theorem to see that it's bigger than 1. If you change x to n and restrict n to the integers, you find that when n \ge 1,$$(1+1/n)^n = \sum_{i=0}^n {n \choose i}{1 \over n^i} \ge 1 + n \cdot {1 \over n} = 2 $$(because all terms are positive). So the limit (if it exists) must be at least 2. To go further, fix an integer ... 4$$\frac{x+x^n}{1+x^n}=1+\frac {x-1}{1+x^n}$$The only problem point is x=-1 for which the denominator is 0 for odd values of n Otherwise the limit is found with little effort. 4 You can use Stirling's approximation, here is another solution. Let u_n=\frac{n^n}{(n!)^2}, we have$$ \frac{u_{n+1}}{u_n}=\frac{(n!)^2}{((n+1)!)^2}\frac{(n+1)^{n+1}}{n^n}=\frac{1}{n+1}\left(1+\frac{1}{n}\right)^n $$Since \left(1+\frac{1}{n}\right)^n=e^{n\ln\left(1+\frac{1}{n}\right)}\underset{n\rightarrow +\infty}{\longrightarrow}e, we have that \lim\... 3$$=\left(\lim_{x\to0}\dfrac{1-\sin^2x}x-\dfrac1{\sin x}\right)-\lim\dfrac{e^x-1}{\sin x}$$The second limit converges to 1 For first, either use \sin x\approx x for x\to0 Or use Are all limits solvable without L'Hôpital Rule or Series Expansion to find$$\lim_{x\to0}\left(\dfrac1x-\dfrac1{\sin x}\right)$$3 The limit doesn't exist because the left-hand limit does not equal the right-hand limit$$-\infty=\lim_{x\to1^-}\frac{1}{\log(x)} \neq \lim_{x\to1^+}\frac{1}{\log(x)}=\infty$$if we extended the real number line and added positive and negative infinity then the limit still wouldn't exist because \infty \neq -\infty. 3 There is a proof based entirely on the methods of differential calculus; see this Differentiability of Exponential Functions by Philip M. Anselone and John W. Lee In that paper you will find the following. Theorem 1. Let f (x) = a^x with any a > 1. Then f is differentiable at  x = 0 and f'(0) > 0. Theorem 2. Let f (x) = a^x with any a &... 3 For x>0 we have  \frac{|x|}{x}=1 therefore$$\lim_{x \to \infty} \frac{|x|}{x}=\lim_{x \to \infty}1=1$$3 Using only limits you have:$$f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{h}= \lim_{h \to 0} \frac{a^h-1}{h}\therefore f'(x) = a^x \times f'(0)However, you cannot prove that f'(0) = \ln a without using the property that e^x is its own derivative. If you accept the fact as described in this answer, use the fact that a^x = e^{x \ln a} = f(... 3 Here's a fun, unconventional way to think about it. If we suppose the limit exists, we have \begin{align} \lim\limits_{x\to 0} \frac{e^x-1}{x^2}&=\lim\limits_{x\to 0} \frac{e^x-1}{x}\cdot\frac1x\\ &=\lim\limits_{x\to 0} \frac{e^x-1}{x}\cdot\lim\limits_{x\to 0}\frac1x\\ &=1\cdot\lim\limits_{x\to 0}\frac1x \end{align} But we know that for \lim\... 3 Instead of moving to Stirling's approximation, you can go with an easy method: ratio test. Specifically, you can compute the ratio of x_{n+1} and x_n as \frac{x_{n+1}}{x_{n}} =\frac{(n+1)^{n+1}}{((n+1)!)^2}\frac{(n!)^2}{n^n} =\frac1{n+1}\left(1+\frac1n\right)^n \to\frac{e}{n+1} \to 0. $$Then the series x_n=\frac{n^n}{(n!)^2} converges. Denote its ... 3$$\lim_{x\to\pi/4}\cot x^{\cot4x}=\left(\lim_{x\to\pi/4}(1+\cot x-1)^{1/(\cot x-1)}\right)^{\lim_{x\to\pi/4}{\cot4x(\cot x-1)}}$$The inner limit converges to e For the exponent,$$\lim_{x\to\pi/4}\cot4x(\cot x-1)=\lim_{x\to\pi/4}\dfrac{\cos4x}{\sin x}\cdot\lim_{x\to\pi/4}\dfrac{\cos x-\sin x}{\sin4x}$$Now$$\lim_{x\to\pi/4}\dfrac{\cos4x}{\sin x}=\...

2

You are correct. The fraction is dependent to $\theta$ and the limit does not exist and is not equal to $1$.

2

Hint. Write $$x^{1/x}=e^{\frac1x\log x}.$$ Then note that $$\frac{1}{x}\le \frac{\log x}{x}\le \frac{\sqrt x}{x},$$ as $x\to +\infty.$

2

$$\sqrt[n]{\sum_{k=0}^{n}{2^{n-k}\cdot 3^k}}=\sqrt[n]{3^{n+1}-2^{n+1}} \overset{n\to \infty}\longrightarrow 3$$

2

You may consider that $$\zeta(s)=\sum_{n\geq 1}\frac{1}{n^s} = \sum_{n\geq 1}\frac{1}{\Gamma(s)}\int_{0}^{+\infty}x^{s-1}e^{-nx}\,dx =\frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x-1}\,dx$$ holds for any $s$ such that $\Re(s)>1$. Similarly $$\eta(s) = \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^s} = \frac{1}{\Gamma(s)}\int_{0}^{+\infty}\frac{x^{s-1}}{e^x+... 2 Assuming that you know that (2^x)'=\log(2)\times2^x, then, in particular, you know that$$\lim_{x\to0}\frac{2^x-1}x=\log(2).$$Therefore,$$\lim_{x\to0}\frac x{2^x-1}=\frac1{\log(2)}.$$2 It’s not directly using \ln in the limit itself Consider$$f’(x) = a^x(f’(0))\frac{f’(x)}{f(x)} = f’(0)$$Taking definite integral$$\displaystyle\int_0^1 \frac{df(x)}{f(x)dx} dx = f’(0)f’(0) = \ln a$$so we have the desired result. 2 Since 1 + x \le e^x for all x and we have e^x \le 1/(1 - x) for x < 1 so$$1 \le (e^x - 1)/x \le 1/(1 - x) \to 1$$as x\to 0. 2 0 \leq f _n(x) \leq \frac 1 {n^{2}}. By Squeeze Theorem \lim_{n\to \infty} f_n(x)=0 for any real number x. 2 Let (1+1/x)^x \to a. Taking logarithm of both sides,$$ \log a = x\log\left(1+\frac{1}{x}\right) = x \left(\frac{1}{x}- \frac{1}{2x^2} + \cdots\right) =\left( 1 - \frac{1}{2x} + \cdots\cdots\right) \to 1 $$If the limit was 1 then we should have got \log a \to 0 which is clearly not the case. 2 What happens when you multiply the numerator and denominator by the (clearly non-zero) conjugate$$\sqrt{(1-h)^2+(2+h)^2} + \sqrt{5}?$$1 Let \varepsilon > 0 be given. Let N > 0. Then we have that \forall x >N:$$\left| \frac{|x|}{x}-1\right|=|1-1|=0<\varepsilon$$So the limit at +\infty is 1. 1 Solving limit using @gimusi 's hint: \lim\limits_{x \to 0}{({\frac{\cos x}{\cos 2x}})^{\frac{1}{x^2}}}=\lim\limits_{x \to 0}{e^{\frac{\ln ({\frac{\cos x}{\cos 2x}})}{x^2}}}=\lim\limits_{x \to 0}{e^{\frac{\ln (\cos x)-\ln (\cos 2x)}{x^2}}} Since \cos x~1-{\frac{x^2}{2}}, So, \lim\limits_{x \to 0}{({\frac{\cos x}{\cos 2x}})^{\frac{1}{x^2}}}=e^{\lim\... 1 Assuming the following limits as known \lim_{y\to 0}(1+y)^{\frac{1}{y}} = e \lim_{x\to 0}\frac{1-\cos x}{x^2} = \frac{1}{2} and \lim_{x\to 0}\frac{\sin x}{x} = 1 you can proceed as follows: Write \frac{\cos x}{\cos 2x} = 1 + \left(\frac{\cos x}{\cos 2x} -1\right) Set y(x) = \frac{\cos x}{\cos 2x}-1 \stackrel{x\to 0}{\longrightarrow}0 Now, note ... 1 Hint$$A={\left({\frac{\cos (x)}{\cos (2x)}}\right)^{\frac{1}{x^2}}}\implies \log(A)={\frac{1}{x^2}}\log\left({\frac{\cos (x)}{\cos (2x)}}\right) Use the series of $\cos(x)$ and $\cos(2x)$ and then long division. Continue with Taylor expansion fo the result. Divide by $x^2$. At this point, you have the Taylor series for $\log(A)$. Continue with Taylor ...

1

\$\begin{array}\\ \dfrac{\cos^2x}{x}-\dfrac{e^x}{\sin x} &=\dfrac{1-\sin^2x}{x}-\dfrac{e^x}{x+O(x^3)}\\ &=\dfrac{1-(x+O(x^3))^2}{x}-\dfrac{1+x+O(x^2)}{x+O(x^3)}\\ &=\dfrac{1-x^2+O(x^3)}{x}-\dfrac{1+x+O(x^2)}{x}(1+O(x^2))\\ &=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)(1+O(x^2))}{x}\\ &=\dfrac{1-x^2+O(x^3)-(1+x+O(x^2)}{x}\\ &=\dfrac{-x-x^2+O(x^3)}{...

Only top voted, non community-wiki answers of a minimum length are eligible