Articles of limits without lhopital

limit without L'Hôpitale rule or infinity series

I was solving this limit $$ \lim_{x\to0} \frac{\ln(x+1)-x}{1-\cos(x)}=L $$ I tried to rewrite the function $f(x)=\dfrac{\ln(xe^{-x}+e^{-x})}{2\sin^2(\frac{x}{2})}$ and we have $$\lim_{x\to0} f(-x)=L$$ $$ 4L=\lim_{x\to0} \frac{(e^x-1)^2}{\sin^2(\frac{x}{2})}=4 $$ and that give us $L=1$ But when I calculate it by Wolfarm the result was $-1$ Can any one tell me where’s my fault ?

Is showing $\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists the same as $\lim_{n \to \infty} (1+1/n)^n$ exists

My expanded question: Is showing $\lim_{z \to \infty} (1+\frac{1}{z})^z$ exists as $z$ goes through real values the same as $\lim_{n \to \infty} (1+\frac{1}{n})^n$ exists as $n$ goes through integer values? If not, how much additional work is needed to make the two equivalent? I am asking this because I had posted a question which stated […]

How to calculate limit without L'Hopital

How can I calculate limit without using L’Hopital rule? $\displaystyle\lim_{x\to 0 }\frac{e^{\arctan(x)}-e^{\arcsin(x)}}{1-\cos^3(x)}$.

How to calculate $\lim\limits_{{\rho}\rightarrow 0^+}\frac{\log{(1-(a^{-\rho}+b^{-\rho}-(ab)^{-\rho}))}}{\log{\rho}} $ with $a>1$ and $b>1$?

How to calculate this question? $$\lim\limits_{{\rho}\rightarrow 0^+}\frac{\log{(1-(a^{-\rho}+b^{-\rho}-(ab)^{-\rho}))}}{\log{\rho}} ,$$ where $a>1$ and $b>1$. Thank you everyone.

Finding $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without L'Hopital and series expansion.

So we have to find $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without using any series expansions or the L’Hopital’s rule. I did it using both but I have no idea how to do it. I tried many substitutions but nothing worked. Please point me in the right direction.

Find $\lim_{x \to 0} x \cdot \sin{\frac{1}{x}}\cos{\frac{1}{x}}$

$$\lim_{x \to 0} x \cdot \sin{\frac{1}{x}}\cos{\frac{1}{x}}$$ I don’t solve this kind of limits, I can’t try anything because it seems difficult to me.

Limit $\lim_{x\to 0} \frac{\tan ^3 x – \sin ^3 x}{x^5}$ without l'Hôpital's rule.

I need to solve $$\lim_{x\to 0} \dfrac{\tan ^3 x – \sin ^3 x}{x^5}$$ I did like this: $\lim \limits_{x\to 0} \dfrac{\tan ^3 x – \sin ^3 x}{x^5} = \lim \limits_{x\to 0} \dfrac{\tan ^3 x}{x^5} – \dfrac{\sin ^3 x}{x^5}$ $=\dfrac 1{x^2} – \dfrac 1{x^2} =0$ But it’s wrong. Where I have gone wrong and how to […]

$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$

Calculate: $$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$ I don’t know how to use L’Hôpital’s Rule. I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}$.

Calculate $\lim_{n \to \infty}\binom{2n}{n}$ without using L'Hôpital's rule.

Questions: (1) Calculate $$\lim_{n \to \infty}\binom{2n}{n}$$ (2) Calculate $$\lim_{n \to \infty}\binom{2n}{n} 2^{-n}$$ without using L’Hôpital’s rule. Attempted answers: (1) Here I start by using the definition of a binomial: $$\lim_{n \to \infty}\binom{2n}{n} = \lim_{n \to \infty} \frac{(2n)!}{n!(2n-n)!} = \lim_{n \to \infty} \frac{(2n)!}{n!n!} = \lim_{n \to \infty} \frac{2n \cdot (2n-1) \cdot (2n-2) \cdot … \cdot (n […]

Difficulty in evaluating a limit: $\lim_{x \to \infty} \frac{e^x}{\left(1+\frac{1}{x}\right)^{x^2}}$

The limit I have to evaluate is this – $$\lim_{x \to \infty} \frac{e^x}{\left(1+\frac{1}{x}\right)^{x^2}}$$ I first checked if L’Hopital’s rule applies here. The limit of both numerator and denominator is $\infty$. But differentiating the denominator yield a even more complicated expression. I am not getting how to approach this question using some other method. Thank you.