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)}{...


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