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.

3046 questions
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Find the limit $\lim_{ x \to \pi }\frac{5e^{\sin 2x}-\frac{\sin 5x}{\pi-x}}{\ln(1+\tan x)}$

Find the limit (without using l'Hôpital and equivalence) $$\lim_{ x \to \pi }\frac{5e^{\sin 2x}-\frac{\sin 5x}{\pi-x}}{\ln(1+\tan x)}=?$$ my try : $u=x-\pi \to 0$ $$\lim_{ x \to \pi }\frac{5e^{\sin 2x}-\frac{\sin 5x}{\pi-x}}{\ln(1+\tan x)}=\lim_{…
Almot1960
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Find the limits without L'Hôpital:$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $

Find the limits without L'Hôpital's rule $$\lim_{ x \to 0 }\frac{x-\sin x}{x-\tan x}=? $$ My Try: $$\lim_{ x \to 0 }\frac{\sin(\pi-x)-\sin x}{\tan(\pi+x)-\tan x}=?\\\lim_{ x \to 0…
Almot1960
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Find: $\lim_{x\to -\infty} \frac{\ln (1+e^x)}{x}$ (no L'Hospital)

Find: $\displaystyle \lim_{x\to -\infty} \dfrac{\ln (1+e^x)}{x}$ (no L'Hospital) I'm getting a hard time solving this limit. The book shows 1 as the answer, Wolfram Alpha shows 0. I could solve easily another problem when the denominator was…
bluemaster
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Solving limit shape $0^0$

I have an exercise which is $\lim_{x \to 0^+} x^x$. I have done it, and the answer is $1$, but I just need to know the easiest way to solve it without l'Hospital's rule, since in my country they don't allow us to use L'Hospital's rule.
Mang Sopheak
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Find $\lim_{x\to e}\frac{x^{e^x}-e^{x^e}}{x-e}$ without L'Hospital

By L'Hospital's rule, it is not hard to find that $$\lim_{x\to e}\frac{x^{e^x}-e^{x^e}}{x-e}=e^{e^e+e-1}.$$ But how to find $\displaystyle\lim_{x\to e}\frac{x^{e^x}-e^{x^e}}{x-e}$ without L'Hospital's rule ? Thank you.
Bless
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How to calculate a limit without using L'Hôpital's Rule

I have to calculate the limits $$\lim_{x\rightarrow 0}\frac{\sin(x\sqrt{x})}{x}\;\;\;\text{and}\;\;\; \lim_{x\rightarrow0}\frac{\sin\sqrt{x}}{x}$$ without using L'Hôpital's Rule, and I can't see how. Calculating those derivatives would help me…
Mouhamad Al Mounayar
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Find the limits of

find the limits : $$\lim_{x\to 0}\frac{\sin 5x-5\sin 3x+10\sin x}{\sin (\sin x)+\tan x-2x}$$ i know that : $$\sin 5x=5\sin x-20\sin^3x+16\sin^5x$$ $$5\sin 3x=15\sin x-20\sin^3x$$ so we have : $$\sin 5x-5\sin 3x+10\sin x=16\sin^5x$$ $$\lim_{x\to…
Almot1960
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How do i find this limit WITHOUT L'hôpitals rule

How do I find $$\lim_{x\to 0}\frac{(x+4)^{3/2}+e^x-9}{x}$$ without l'hôpital rule? I know from l'hôpital the answer is 4.
Umeko Inoue
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$\lim_{ x \to0^- }\frac{2^{\frac{1}{x}}+2^{\frac{-1}{x}}}{3^{\frac{1}{x}}+3^{\frac{-1}{x}}}=?$

fine the limits-without-lhopital rule $$\lim_{ x \to0^- }\frac{2^{\frac{1}{x}}+2^{\frac{-1}{x}}}{3^{\frac{1}{x}}+3^{\frac{-1}{x}}}=?$$ My Try : $h= \frac{1}{x} :h\to - \infty$ so : $$\lim_{ h\to - \infty }\frac{2^{h}+2^{-h}}{3^{h}+3^{-h}}=?\\\lim_{…
Almot1960
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$\lim \limits_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$

find the limit : $$\lim_{x\to 0} \frac {4^{\sin x}-2^{\tan 2x}}{3^{\sin 2x}-9^{\tan x}}=?$$ my try : $$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\tan 2x}}{3^{\sin 2x}-3^{2\tan x}}=$$ $$\lim_{x\to 0} \frac {2^{2\sin x}-2^{\frac{\sin 2x}{\cos2x}}}{3^{\sin…
Almot1960
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Can I prove that $\frac{e^x - e^{\sin x}}{x - \sin x} \to 1$ as $x \to 0$ without using DLH?

Can I prove that $$\frac{e^x - e^{\sin x}}{x - \sin x} \to 1$$ as $x \to 0$ without using DLH? UPDATE: If you apply MVT at $f(u) = e^u$ at $[x,\sin x] \forall x>0$ and at $[x,\sin x] \forall x<0$ and use the squeeze theorem the limit is equal to 1.
bolzano
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limit of $\lim_{x\to 0}\frac{\ln(\tan(x)+1)-\sin(x)}{x\sin(x)}$

I am trying to compute the following limit without L'Hôpital's rule : $$L=\lim_{x\to 0}\frac{\ln(\tan(x)+1)-\sin(x)}{x\sin(x)}$$ I evaluated ths limit using L'Hôpital's and found $-\frac12$ as the answer. I eventually ended up to :…
T.D
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Determining a limit without L'Hospital's Rule

I'm trying to solve the limit $$\lim_{h\to0}\frac{\sin(\pi+h)}{h(\pi+h)}$$ without using L'Hospital's rule. It's part of a problem for which I am trying to prove that $$\lim_{h\to0}\frac{\frac{\sin(\pi+h)}{|\pi+h|}-\frac{h}{\pi}}{|h|}$$ exists.…
k.jackson
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Find the $\lim_{x \to \infty} \Big(\frac{x}{e} - x\Big(\frac{x}{x+1}\Big)^x\Big)$

$$\lim_{x \to \infty} \bigg(\frac{x}{e} - x\Big(\frac{x}{x+1}\Big)^x\bigg)$$ I got the answer to the above limit by applying the L'Hôpital twice by first taking $x$ outside and then dividing it to make the limit of the form $\frac{0}{0}$ I want to…
Tony
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$ \lim_{n \to \infty} \sqrt[n]{a^n+1}$ with $a \ge 0$

I'm a bit rusty with limits $$ \lim_{n \to \infty} \sqrt[n]{a^n+1}$$ with $a \ge 0$. The solution in my book is $max \left \{ 0,1 \right \}$ but my final results are: 1) $+\infty$ if $01$ 3) $+\infty$ if $a=1$
Anne
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