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Questions tagged [limits-without-lhopital]

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

1
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4answers
56 views

Evaluate $\lim\limits_{x\to+\infty} \sqrt{x}\left ( \sqrt[3]{x+1}-\sqrt[3]{x-1}\right )$?

How do you evaluate $$\lim\limits_{x\to+\infty} \sqrt{x}\left (\sqrt[3]{x+1}-\sqrt[3]{x-1}\right ) ?$$ Thanks in advance for your support.
3
votes
7answers
127 views

Computing $\lim \limits_{x\to0}\frac{1}{x}\left[\frac{1}{x}\left(e^x-1\right)-1\right]$ without using l'Hopital [on hold]

$$\lim \limits_{x\to0}\frac{1}{x}\left[\frac{1}{x}\left(e^x-1\right)-1\right]$$ How do I calculate this without using l'Hopital rule?
2
votes
2answers
57 views

Prove $\lim_{(x, y) \to (0, 0)} f(x)$ exists if and only if $m + n > 2$

Problem: Let $m, n \in \mathbb{N}$. Show that $\lim_{(x, y) \to (0, 0)} \frac{x^{m}y^{m}}{x^{2} + y^{2}}$ exists if and only if $m + n > 2$. My try: I'm really not too sure about how to prove ...
0
votes
3answers
24 views

Limit of a function with natural log to a power [on hold]

Evaluate the limit. a, b, and c are all constants. $$\lim_{(t)\rightarrow (0)} \frac{a^\frac43c(1-\ln(e+bt))^\frac53}{t^\frac53(a^4+b^4+3^4c^4)}$$
2
votes
3answers
101 views

How can I solve this crazy limit? $\lim _{x\rightarrow0}\frac{1-\cos\left(\frac{1-\cos x \cos 2x}{x^2}-\frac {5}{2}\right)\cos2x}{x^2} $

Firstly, I think this can be done with equivalent infinitesimal, but it seems so much complicated. I'm not very brave to do L'Hospital's rule on this question. And I don't think trig formulas can ...
1
vote
3answers
54 views

Showing that $\lim_{(x,y,z)\to(0,e,-2/3)}\frac{\sqrt[3]{x^4}\left(1-\ln y\right)^\frac 53\left(z+\frac23\right)}{x^4+(y-e)^4+(3z+2)^4}$ does not exist

Show that the following limit does not exist: Solve using the method described below. $$\lim_{(x,y,z)\rightarrow (0,e,-2/3)} \frac{\sqrt[3]{x^4}\left(1-\ln(y)\right)^\frac 53 \left(z+\frac23\...
2
votes
1answer
24 views

Applying comparison theorem to limits when inequality is known

I somehow confuse myself whenever I apply the comparison lemma. I know there is the comparison lemma that says the following: Let the sequence $\{a_{n}\}$ converge to the number $a$. Then the ...
1
vote
3answers
44 views

What real $p$ makes this limit exists and be finite?

The limit is $\lim_{x \to 0} \frac{\sqrt{1+x}-(1+px)}{x^2}$ We must find what real $p$ makes this limit exist and be finite, and determine this limit. I know it can be done using derivative ideas, ...
2
votes
4answers
70 views

How do I find the limit of $\frac{x \cdot \ln(1+x)}{e^{2x}-1}$

How do I find: $$\lim_{x \to 0} \frac{x \cdot \ln(1+x)}{e^{2x}-1}$$ without using L'Hospital. I know that $$\lim_{x \to c}\frac{f(x)}{g(x)} = \frac{L}{M}, M \neq 0$$ My problem is I get: $M = 0$ ...
1
vote
1answer
29 views

Greatest lower bound of the set : $\{(e^n + 2^n)^\frac1n\ | \; n\in \mathbb{N}\}$

Find the greatest lower bound of the set : $\{(e^n + 2^n)^\frac1n \mid n\in \mathbb{N}\}$ I will find the limit of $a_n := (e^n + 2^n)^\frac1n $ which if exists, must equal the greatest lower ...
0
votes
0answers
43 views

Convergence of $a_n=\sum\limits_{m=2}^{n} \frac{(-1)^mm}{(\ln(m))^m}$ and $b_n = \sum\limits_{m=2}^{n} \frac{1}{(\ln(m))^m}$

Let $a_1=b_1=0$ and for each $n\geq 2$, let $a_n$ and $b_n$ be real numbers given by $a_n=\sum\limits_{m=2}^{n} \frac{(-1)^mm}{(\ln(m))^m}$ and $b_n = \sum\limits_{m=2}^{n} \frac{1}{(\ln(m))^m}$,...
3
votes
1answer
43 views

$(a_n) $ is a sequence of positive real numbers. The series $\sum a_n$ will converge if

$(a_n) $ is a sequence of positive real numbers. The series $\sum a_n$ will converge if (a) $\sum a_n^2$ converges. (b)$\sum \frac{a_n}{2^n}$ converges (c)$\sum \frac{a_{n+1}}{a_n}$ coverges (d)$\...
2
votes
1answer
25 views

Let $(a_n),(b_n) $ be sequences of positive real numbers such that $na_n<b_n<n^2a_n$

Let $(a_n),(b_n) $ be sequences of positive real numbers such that $na_n<b_n<n^2a_n$ for all $n\geq2$. If the radius of convergence of $\sum a_n x^n$ is $4$, then $\sum b_n x^n$ A)converges ...
0
votes
2answers
48 views

Finding $\lim_{{n \to \infty}} \frac{1+\sqrt[4]{2}+\cdots+\sqrt[4]{n}}{n\sqrt[4]{n^3}\left(\sqrt[4]{(n+1)^2+1}-\sqrt[4]{n^2+1}\right)}$ [on hold]

Find $$\lim_{{n \to \infty}} \frac{1+\sqrt[4]{2}+\cdots+\sqrt[4]{n}}{n\sqrt[4]{n^3}\left(\sqrt[4]{(n+1)^2+1}-\sqrt[4]{n^2+1}\right)}$$ Any ideas other than using Stolz-Cesaro?
0
votes
2answers
33 views

Method for finding a sequence's limit

i'm trying to find the following limit, if it exists, $$\lim_{n→ ∞} \frac{(n+7)^{n-5}}{n^n}$$. Now, I've tried division like $$\lim_{n→ ∞}\frac{|n+1|}{|n|}$$, or dividing by the highest power, but I ...
1
vote
1answer
38 views

Evaluating limit with 2 unknown parameters

I have another question involving limits with unknown parameters: In this one, the limit approaches infinity: $\lim \limits_{x \to 1 } {{\frac{x^2+ax+b}{(x-1)^2}}}= \infty $ This one is really ...
0
votes
3answers
44 views

Find $a$ such that $\lim \limits_{x \to 1 } {{\frac{x^2+(3-a)x+3a}{(x-1)}}}= 7$

I have this limit equation below: $$ \lim \limits_{x \to 1 } {{\frac{x^2+(3-a)x+3a}{(x-1)}}}= 7 $$ One needs to determine the value of $a$ (where $a$ is a real number) such that as $x$ approaches $1$...
5
votes
1answer
42 views

Limit of $\lim \limits_{x \to\infty} (\frac{x}{1-x})^{2x}$

I've been struggling with this term for a while now: $$\lim \limits_{x \to\infty} (\frac{x}{1-x})^{2x}$$ I know it has to do something with $\lim \limits_{x \to\infty} (1+\frac{n}{x})^x=e^n$ but didn'...
0
votes
4answers
62 views

How to evaluate the limit without using L'Hopital? [closed]

$$\lim_{x\rightarrow \pi}\frac{\cos x+\sin 2x+1}{x^2-\pi^2}$$ After using the L'Hopital, I get it is $\frac{1}{\pi}$, but how to evaluate it without using the rules? The note on the paper is: $\...
1
vote
2answers
51 views

Find the limit as $(x,y)$ approaches $(0,0)$ of $\frac{\sin(x)(1-\cos(y))}{6xy^{2}}$

I used the path $y=x$ and got that the limit equals $12$. How do I manipulate the original expression to prove that the limit actually equals $12$ for all paths? Thank you!
2
votes
6answers
502 views

How to calculate this limit without L'Hopital's rule

I know that in the very small values of $x$ $$(1+x)^n ≈1+ nx$$ and I can prove it using Taylor series. But I wanted to prove it without any smell of derivative. So... In order to calculate the ...
1
vote
3answers
48 views

Find the limit without L'Hôpital, fraction with cube root?

How can i find this limit without using L'Hôpital (or anything using derivatives for that matter)? $$\lim_{x \to -8} \frac{\sqrt{1-x}-3}{2+\sqrt[3]{x}} $$ I have tried various substitutions and ...
-2
votes
4answers
98 views

Proof that $\lim_{x \to 0} \frac{\sin(x) - x}{x^2} = 0$ (Without L'Hospital) [duplicate]

I was studying calculus and I got stuck in proving that $$\lim_{x \to 0} \frac{\sin(x) - x}{x^2} = 0.$$ Using L'Hospital is easy. However, I want a proof where I don't use L'Hospital. Help?
0
votes
0answers
52 views

$\lim _{x\to 0}\left(\sin\left(x\right)\cos\left(\frac{5}{x}\right)\right)$ Can I apply Sandwich Theorem to this limit?

Actually the original question was the following Determine whether the following limit exist? $\lim _{x\to 0}\left(x^2\sin\left(\frac{2}{x}\right)+\sin\left(x\right)\cos\left(\frac{5}{x}\right)\...
3
votes
2answers
49 views

$\lim_{x\to0}\frac{\arcsin x-\sin x}{x^3}$ without series or L'Hospital

$$\lim_{x\to0}\frac{\arcsin x-\sin x}{x^3}$$ without using series or L'Hospital Is there any ohter simpler method? Expansion of $\arcsin$ is not trivial like tha of sine and L'Hospital is too ...
0
votes
1answer
31 views

help with finding that fx =gx

I am trying to multiply fx with the given equation to get gx but im not getting the right answer. any help would be appreciated. The link to the question is posted. Define functions $f(x)$ and $g(...
0
votes
2answers
37 views

how to define the functions $f(x)$ and $g(x)$ by a given equation

Define functions $f ( x )$ and $g ( x )$ by $f ( x ) = \frac { \sqrt [ 3 ] { x } - 2 } { x - 8 } , \quad g ( x ) = \frac { 1 } { \sqrt [ 3 ] { x } ^ { 2 } + 2 \sqrt [ 3 ] { x } + 4 }$ (a) By ...
2
votes
3answers
59 views

Calculate limit with squares $\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4$

$$\lim_{n \to \infty}\left(\sqrt[3]{n^3+2n-1}-\sqrt[3]{n^3+2n-3}\right)^6 \cdot (1+3n+2n^3)^4 $$ What I did was to multiply it and I got $\frac{1}{2}$ as the final result. Could someone confirm if it'...
2
votes
3answers
57 views

Solving limit without applying L'Hospital's rule

There's a practice problem I'm utterly stumped at. No matter what I try, I can't arrive at the correct answer. Please, if someone could help me understand, I would be very grateful: $$ \lim_{x\to \...
0
votes
4answers
71 views

How to compute $\lim_{x\to \infty} \frac{\ln(1+x)}{\ln x} $?

How to compute $\lim_{x\to \infty} \frac{\ln(1+x)}{\ln x} $? I managed to find it is $1$ by using L'Hôpital's, but I am sure there is also an elementary solution.
0
votes
2answers
55 views

Radius of convergence of power series $\sum_{n=0}^{\infty} n!x^{n^2}$

Radius of convergence of power series $\sum_{n=0}^{\infty} n!x^{n^2}$ $\sum_{n=0}^{\infty} n!x^{n^2} = 1 + x + 2x^4 + 6x^9\ldots$ Comparing this with $\sum_{n=0}^{\infty} a_nx^n=$ $a_n= n! $ or $0$ ...
5
votes
2answers
101 views

help with solving a limit with logarithms

I am preparing for an exam and I am struggling with the following limit: $$ \lim_{x\to 0}\frac{\ln(1+x^{2018})-\ln^{2018}(1+x)}{x^{2019}}. $$ I tried the L'Hospital rule and i tried to form ...
2
votes
1answer
53 views

$a_n=(-1)^{n-1}, \; s_n=\sum_{i=1}^{n}a_i$ then find $ \lim_{n\to \infty}\frac{s_1+s_2+\dots s_n}{n}$

$a_n=(-1)^{n-1}, \; s_n=\sum_{i=1}^{n}a_i$ then find $ \lim_{n\to > \infty}\frac{s_1+s_2+\dots s_n}{n}$ $$s_k=1,\; \text{if k is odd and } s_k=0 \text{ if k is even} $$ Cauchy's theorem for a ...
1
vote
1answer
62 views

Two different answers with different methods in solving limits.

So the question is, $$\lim_{x\to 0}\{\frac{(1+x){\log(1+x)} - x}{x^2}\}$$ Method 1: Applying l'hopital's rule the limit tend to $\bigl(\frac{1}{2}\bigr)$. Method 2: Solving the following way: $$\...
0
votes
1answer
47 views

Finding limits of $f: \mathbb{R}^n \to \mathbb{R}$ (at the origin) for general $n$

Recently, I have learned that $$\lim_{(x,y) \to(0,0)} \frac{x-y}{ \sqrt{x}- \sqrt{y}}=0$$ from here: https://math.stackexchange.com/a/2202266/638440 . Let $f: \mathbb{R}^3 \to \mathbb{R}$ be a ...
1
vote
1answer
53 views

Directional derivative of $f(x,y ) = \frac{xy^2}{x^2+y^4}, (x,y)\neq (0,0)$ & otherwise $0$

Find the directional derivative of $$ f(x,y) := \begin{cases} \frac{xy^2}{x^2+y^4}, & \text{for } (x,y)\neq (0,0) \newline 0, & \text{otherwise.} \end{cases} $$ at $(0,0)$ in direction of ...
1
vote
1answer
37 views

Continuity at $(0,0)$ of $f(x,y)=\frac{x^3+y^3}{x-y}, \; \;(x,y)\neq(0,0)$, otherwise $0$

$f(x,y)=\frac{x^3+y^3}{x-y}, \; \;(x,y)\neq(0,0)$, otherwise $0$ If I use $x=r\cos(t), \; y=r\sin(t)$ & $r\to 0$, I get the limit equal to $0$. Is this sufficient to prove that the limit of the ...
0
votes
0answers
19 views

L'hopital rule for vector sequences?

I need to find the limit $$ \lim_{r\to 0}{\frac{B.r}{a^*.r}}, $$ where $B$ is a $n\times n$ matrix in $C^{n\times n}$, $a$ and $r$ are vector columns in $C^n$. Vector $a^*$ is the conjugate transpose ...
4
votes
3answers
1k views

How can I claim a one-sided limit doesn't exist?

I have to find the limit $$\lim_{x\to 0^+} \frac{\ln(1+2x)\sin x}{\sqrt {x^3}} $$ Now, I tried using L'Hôpital's rule, but it doesn't lead anywhere. Manually trying to convert the functions to ...
1
vote
6answers
138 views

Evaluate ${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$

Evaluate $${\lim_{x \rightarrow 0} \frac{1-\cos x\cos 2x\cdots \cos nx}{x^2}}$$ It should be $$\frac{1}{12}n(n+1)(2n+1)$$ but I don't know how to prove that. I am also aware that $\lim\limits_{\theta ...
2
votes
3answers
50 views

Calculating the Limit of a Function without use of L'Hospital's Rule or Power Series

I came across this question the other day and have been trying to solve it by using some simple algebraic manipulation without really delving into L'Hospital's Rule or the Power Series as I have just ...
3
votes
3answers
57 views

How to find limit $\lim_{h\rightarrow0^-}\frac{e^{-1/|h|}}{h^2}$

How can I find limit $$\lim_{h\rightarrow0^-}\frac{e^{-1/|h|}}{h^2}$$ I solve subproblem: $$\lim_{h\rightarrow0^-}\frac{e^{-1/|h|}}{h} = \lim_{h\rightarrow0^-}\frac{1}{e^{1/|h|}\cdot h} =\lim_{y\...
1
vote
2answers
39 views

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$

How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}\frac{\cos(ah)a}{1} = a$$ but I don't know how to behave ...
1
vote
1answer
48 views

Different results with different methods to same limit of polynomial

I'm considering the following limit while looking for asymptotes: $$\lim_{x\to \infty} \arctan\dfrac{\sqrt{x^2+1}}{x-1}$$ Going the route I initially tried, I divide both parts of the fraction by $x^...
0
votes
3answers
44 views

Trigonometric limits [closed]

Please help me to find the value of $$ \lim_{x→1}⁡\frac{1+\cos(\pi x)}{\tan^2(\pi x)} $$ which is in zero over zero form without using L’Hôpital’s rule.
2
votes
3answers
61 views

Why I can substituto $\cos x$ in this limit but no $e^x$?

$$\lim_{x\to 0} \dfrac{(e^x-\cos x)^5}{x^4·\sin x ·\ln (2+x)}=\frac{1}{\ln 2}$$ I know how to solve that limit using equivalences and I get that answer, however it gave me a lot of doubts. Like I ...
1
vote
3answers
57 views

Evaluate the limit: $\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9}$

$$\lim_{x\to 3} \frac{x^{2}+\sqrt{x+6}-12}{x^{2}-9} $$ I want to know how to evaluate without using L'Hopital Rule. I'm unable to factorise or simplify it suitably.
1
vote
4answers
61 views

Idea for $\lim\limits_{x \to \frac{\pi}{2}} \left( \tan \left( \tfrac{\pi}{4} \sin x\right)\right)^{1/ ( \tan (\pi \sin x))}$

$$\lim_{x \to \frac{\pi}{2}} \tan \left(\frac{\pi}{4}\sin x\right)^\left({\dfrac 1{\tan(\pi \sin x)}}\right).$$ Need help for this. I tried using trigonometric identities but nothing seems to fit. I'...
5
votes
2answers
90 views

How to evaluate: $\lim\limits_{n\to\infty} \frac{1^{p-1}+2^{p-1}+…+n^{p-1}}{n^p}$

How to evaluate: $$\lim_{n\to\infty} \dfrac{1^{p-1}+2^{p-1}+...+n^{p-1}}{n^p}$$ when $i)$ $p\in\mathbb R,p\neq0$ $ii)\space p=0$ So for $i)$ I tried using Stolz–Cesàro theorem and Binomial ...
0
votes
1answer
62 views

I need to calculate a limit using Euler number [closed]

How would one evaluate this limit without using L'Hôpital's rule? What I know for sure is that this limit equals to zero, but I don’t know how to solve it. $$ \lim_{x\rightarrow \infty}\left(\frac{2x-...