# Questions tagged [limits-without-lhopital]

The evaluation of limits without the usage of L'Hôpital's rule.

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### Answer for the $\lim_{x\rightarrow 0} \frac{x^2}{1-\cos x}$?

So I know that you can multiply by the conjugate and get the correct answer which is 2. However I wanted to know why this method gave me the wrong answer. how I tried to solve it
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### Proving $\lim_{x\to0}\sin\left(\frac{\pi}{x}\right)$ is undefined using the $\epsilon$-$\delta$ definition of a limit

It is well known that $$\lim_{x\to0}\sin\left(\frac{\pi}{x}\right)$$ is undefined, which is intuitively true since the function is periodic and oscillates between $1$ and $-1$ as $x$ approaches zero, ...
1 vote
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### Why is $\delta = \epsilon$ for limit of $f(x)=x$ at any point?

[This question is rather a very easy one which I found to be a little bit tough for me to grasp. If there is any other question that has been asked earlier which addresses the same topic then kindly ...
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### Solving the limit $\lim_n \left(1+\frac{1}{-n} \right)^{-n}$

I was doing an exercise about limits of sequences and arrived at the following limit: $$\lim_n \left(1+\frac{1}{-n} \right)^{-n}\ \ \ \ (1)$$ We are supposed to solve the limit without using L'hopital'...
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### Evaluating $\lim_{x\to {0}}\frac{1}{x\arcsin x} - \frac{1}{x^2}$ without L'Hôpital's rule

So I have this limit ... $$\lim_{x\to {0}}\frac{1}{x\arcsin x} - \frac{1}{x^2}$$ Using l'hôpital rule, I know the answer is $-\frac{1}{6}$, but it seems like my professor want me to find another way ...
1 vote
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### Is there a limit which is hard to compute without L'Hôpital's rule?

I know that for limits having $0/0$ or $\infty/\infty$ form, L'Hôpital's rule is a great tool. But usually these problems can be solved without using it like using Taylor series, etc. So I wanted to ...
1 vote
62 views

### Can we use $(1+x)^n = 1+nx$ where $x\to0$ and $n$ is $1/0$?

I was solving a limits questions: $$\lim_{x\toπ/4} \tan x^{\tan2x}$$ After putting $x = (π/4)+h$ and solving it I got the expression: $$(1-h/1+h)^{\cot2x} = (1-2h)^{\cot2h}$$ $$(1-2h)^{\cot2h}$$ Now ...
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### $L_1=\lim_{x\to 0}\dfrac{1-\cos x\cos 2x \cos 3x}{x^2}\;$ and $L_2=\lim_{x\to 0} \dfrac{1-(\cos x)^{{(\cos 2x)}^{(\cos 3x)}}}{x^2}\;$

If $$L_1=\lim_{x\to 0}\dfrac{1-\cos x\cos 2x \cos 3x}{x^2}\;$$ and $$L_2=\lim_{x\to 0} \dfrac{1-(\cos x)^{{(\cos 2x)}^{(\cos 3x)}}}{x^2},\;$$ then value of $|L_1-L_2|$ is equal to: My Approach: I put ...
1 vote
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### Hint for proving that $\lim_{x \to 0} \log_{10}{|x|}$ does not exist

As the title suggest, i am currently working on an exercise which asks me to prove that $$\lim_{x \to 0} \log_{10}{|x|}$$ does not exist. The proof is via contradiction. My approach so far has been ...
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1 vote
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### Limit of sin(sin(sin(x)))

I had an exam with the exercise $$\lim _{x\to 0}\left(\frac{\sin(\sin(\sin(x)))}{x}\right)$$ but I needed to solve it without using L'hopital rule but I was not sure how to solve it, do you know how ...
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### Evaluate the common limit [closed]

I'd like more of an explanation than a solution, I'm sorry but I'm studying math again after 12 years, and I don't understand basic concepts. I have to evaluate some limits without using L'Hopital ...
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### Limits involving exponents

I don't understand this statement from Wolfram Alpha: Since $5^{2k+1}$ grows asymptotically slower than $3^{4k+1}$ as $k$ approaches $\infty$, $$\lim_{k\to\infty} 3^{-4k-1}\cdot 5^{2k+1} = 0.$$ ...