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 : $\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$

Find the limit: Without the use of the L'Hôspital's Rule $$\lim_{ x \to 1}\frac{\sqrt[n]{x^n-1}}{\sqrt[n]{nx}-\sqrt[n]{n}-\sqrt[n]{nx-n}}$$ My try: $u=x-1$ Now: $$\lim_{ x \to…
Almot1960
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Solve $\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$

How do I solve for $A$? $$\lim_{x\to 0} \frac{\sin(2x)-\sin(Ax)}{x+x^3} = A^2$$ Since the denominator evaluates to $0$, I tried doing $$\lim_{x\to 0} [\sin(2x)-\sin(Ax)]=A^2 \cdot \lim_{x\to0}[x+x^3]$$ but it would go into $0=0$. If I checked from…
Gyakenji
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Finding limit of $f(x)$ using appropriate method

Consider $f(x) = \frac{1}{1+ \cos(x)}$ . I want to find $\lim _{x\to \infty} f(x)$ using appropriate method. My try : I'm really confused about limit of $\sin(x)$ and $\cos(x)$. Therefore I can't solve this problem.
S.H.W
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How to calculate $\lim\limits_{x \to 0 } \frac{x}{\sin{\pi(x+2)}}$

I divided the limit by the product of the two limits. The first limit is found, but how to calculate second: $\lim\limits_{x \to 0 } \frac{x}{\sin{\pi(x+2)}}$
Dave
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Limit $\lim_{x\to 0} \frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}$

I know that: $$\lim_{a\to 0} \frac{\ln(1+a)}{a}=1$$ But why $$\lim_{x\to 0} \frac{-\frac{2x}{e}}{\ln(1+(-\frac{2x}{e}))}$$ is 1 too? $$\lim_{x\to 0} 1 : \frac{\ln(1+(-\frac{2x}{e}))}{-\frac{2x}{e}}=1:\lim_{x\to 0} …
Dave
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How to solve this limit ( No l'Hôpitale)

Hi I'm sorry for bothering you a lot, but i'm practising for my last test When ever i try to do something it's inverting to NaN situation Can any one help me Thanks a lot Note : Obviously It's not 2sided limit (0+)(0-) With L'Hôpitale I've…
AbdulQader Qassab
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Solving $\lim_{x\to0}\frac{ 5^x-3^x}{x}$ without L'Hôpital's rule

The question is $$\lim_{x\to0}\frac{ 5^x-3^x}{x}$$ Unfortunately, we cannot use L'Hôpital's rule to solve this limit, so I'm not sure how to proceed with the question. I'm sure there is an elegant way to manipulate the algebra to resolve the…
JennyS95
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limit of $\frac{xy-2y}{x^2+y^2-4x+4}$ as $(x,y)$ tends to $(2,0)$

Am I able to substitute $x$ by ($k$+$2$) with $k$ tending to $0$, then using polar coordinates to deduce its limit?! \begin{equation*} \lim_{(x,y)\to(2, 0)}{xy-2y\over x^2+y^2-4x+4}. \end{equation*}
mandez
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What is infinite limit of rational trigonometric function?

How can we justify the following limit? $$ \lim _{x \rightarrow -\infty} \dfrac{1+\cos x}{3+2\cos x} $$ We can see the result of limit DNE or $[0, 2]$ on the graph. How can we show?
Ramazan Özarslan
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Evaluate this limit without LHopital

Using definition of limits, need to find $g'(x)$ for $g(x)= \dfrac{\sin x}{15}$ $g'(x) = \lim_\limits{h \to 0} \dfrac{g(c+h) -g(c)}{h} = \lim_\limits{h \to 0} \dfrac{\dfrac{\sin (x+h) - \sin x}{15}}{h} = \lim\limits_{h \to 0} \dfrac{ \sin x(\cos…
user307640
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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
Yinka Eromomene
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Calculate $\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x$

Calculate $\lim_{x\to \infty}(\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}})^x$. First off, it's easy to see that $\lim_{x\to \infty}\frac{\sqrt[x]{2} + \sqrt[x]{3}} {\sqrt[x]{4} + \sqrt[x]{5}}$ = 1. Therefore, I tried the…
RufP
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Solve $\lim_{x\to 1}\frac{\sin(x^2-1)}{x-1}$ without L’Hopital.

$$\lim_{x\to 1}\frac{\sin(x^2-1)}{x-1}$$ I have tried using squeeze theorem but that doesn’t work. I know you can use derivatives to find that it equals two but is there another to do it?
und
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Hint on finding the limit of this function

If $p(x)$ is a polynomial of degree $10^{2011}$, then: $$\lim_\limits{x \to \infty} p(x)e^{x} = ?$$ How do I even begin with this? I tried expressing it into a form where L'Hospital rule is applicable, but found that it cannot be expressed as such.…
Natasha J
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$\lim_{x\to\pi/2} \cos x\cdot\cos(\tan x)$

I wanted to solve the problem without L'Hopital's rule and was having problems in how to show that the limit does not exist. Any pointers on how I could approach the problem would be super helpful.
Tanay Nagar
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