Mathematics Stack Exchange
Mathematics Stack Exchange

Questions tagged [limits-without-lhopital]

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

The idea here is to evaluate the limit using standard limit theorems (algebra of limits, Sandwich/Squeeze Theorem, essentially without using any differentiation) and some standard limit formulas related to algebraic, trigonometric, exponential and logarithmic functions. Very often, Taylor series techniques prove fruitful in such problems as they allow for easy cancellation of powers and most terms evaluate to zero, leaving a simple expression for the limit.

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$\lim_{ x \to\frac\pi4 } \frac{1-\tan(x)}{\sin(x)-\cos(x)}$ without L'Hospital rule

I have to compute this limit $$\lim_{ x \to\frac\pi4 } \frac{1-\tan(x)}{\sin(x)-\cos(x)}=\lim_{ x \to \frac\pi4 } -\frac{1}{ \cos x }=-\sqrt{2}$$ Is it right? In my book the suggested solution is $-\frac{1}{ \sqrt{2} }$
Anne
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determinination of limit of trigonometric sequence

The value of $$\lim_{n\to\infty}\left(\sin\frac{\pi}{2n}\cdot \sin\frac{2\pi}{2n}\cdot \sin\frac{3\pi}{2n}\cdots \sin\frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}}$$ is equal to? Can anyone remind me of any formula for such series involving sine and…
shadow kh
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Compute specific lim without l'hopital rule

how can i compute this limit without L'hopital rule: $\lim \limits _{z\to 0} \frac{1-cos z}{z^2}$ i believe the answer is 1/2 by the rule, thanks
user406656
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Is my limit development correct?

I have this limit to find: $\lim\limits_{x\to0}\frac{\sqrt[n]{1+x}-1}{x}$ My development was: Let $\large{u^n = 1+x}$, from here if $x\to 0$ implies that $\large{u^n \to 1}$ And i got: $\Large{\lim_{u^n \to 1}\frac{u-1}{u^n - 1}}$ and using that…
ESCM
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How can I prove that $\mathcal L(t)$ equals $s^{-2}$?

I'm trying to prove that $$\mathcal{L}(t) = \frac{1}{s^2}$$ from the definition. After using integration by parts, I got: $$\mathcal{L}(t)=\lim_{t \to \infty}\left (\frac{-t}{s} * e^{-st}-\frac{1}{s^2} * e^{-st}\right) - \lim_{t \to 0}…
U.AL
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$\lim\limits_{x \to +\infty} \frac{\cos(2x^{-\frac{1}{2}})-1}{\frac{\pi}{2}-\arctan(3x)}$

I'm having difficulties with this limit. We haven't covered integrals, derivatives, or things such as L'Hopital's rule in our course yet. What is the correct process to go about this limit? $$\lim\limits_{x \to +\infty}…
Samuele B.
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Trouble solving this limit without using l'Hôpital's rule

The limit I want to calculate is the following $$ \lim_{x \to 0}{\frac{(e^{\sin(4x)}-1)}{\ln\big(1+\tan(2x)\big)}} $$ I've been stuck on this limit for a while and I don't know how to solve it please help me.
Levio
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Alternate solution to a limit without using L'Hopital's rule

$$\lim \limits_{x \to a} \frac{x}{x-a} \left(\frac{x^3}{(a-1)^2}-\frac{a^3}{(x-1)^2}\right)$$ I've gotten to this $$a \cdot \lim \limits_{x \to a} \frac{x^5-2x^4+x^3-a^5+2a^4-a^3}{(x-a)(a-1)^2(x-1)^2}$$ since as far as I'm concerned it's just a…
Ariel Arévalo
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For $\alpha>1$ find $\lim_{n\to\infty} \sqrt[n]{\sqrt[2^n]{\alpha}-1}$

For $\alpha>1$ find (w/o lhopital)$$ \lim_{n\to\infty} \sqrt[n]{\sqrt[2^n]{\alpha}-1} $$ Progress: Used this:$$ x^n - y^n = (x-y)(x^{n-1} + x^{n-2} y + ... + x y^{n-2} + y^{n-1}) $$ to get to this:…
Slavik Egorov
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Evaluate $\lim_{x\to 0^+}x^{\sin x}$ without L’Hospital

$$\lim_{x\to 0^+}x^{\sin x}$$ I know it can be easily solved using L’Hospital’s Theorem, but I would like to see another way. Thanks in advance
Lin Tanaka
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$\lim_{x \to -\infty}\left(1+ \frac{1}{x}\right)^x$

I'm a bit rusty with limits but my book says that $$ \lim_{x \to -\infty}\left(1+ \frac{1}{x}\right)^x= e$$ and I don't agree completely, knowing that $$ \lim_{x \to +\infty}\left(1+ \frac{1}{x}\right)^x=e$$ In my opinion the right result should be…
Anne
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How to find $\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$?

How to find $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{x+x^2-2}$$ My Try : $$x^2+x-2=(x-1)(x+2)$$ $$\lim_{x \to 1} \frac{\sqrt{x}+\sqrt{x+3}-3}{(x-1)(x+2)}\cdot \frac{(\sqrt{x}+\sqrt{x+3})+3}{(\sqrt{x}+\sqrt{x+3})+3}=\\ =\lim_{x \to 1}…
Almot1960
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I need help solving a trigonometric limit.

I'm seriously stuck. The limit is $(x-3)\,\csc(\pi x)$ where $x$ approaches $3$. Also, please don't make use of l'Hôpital's rule. Thanks in advance.
Alvaro Morales
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Finding limit of $\lfloor x \rfloor + \lfloor -x \rfloor $

Consider $f(x) = \lfloor x \rfloor + \lfloor -x \rfloor $ . Now find value of $\lim_{x \to \infty} f(x) $ . I know that if $x_0 \in \mathbb{R}$ then $\lim_{x \to x_0} f(x) = -1$ but I don't know whether it is true or not in the infinity .
S.H.W
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$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$

if : $$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$ then $a+b=?$ Without the use of the L'Hôspital's Rule My Try : $$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$ $$\lim_{ x \to 0 }\left(…
Almot1960
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