I want to solve this limit without the use of L'Hôpital's rule:  $$\lim_{x\to 0} \frac{e^x-1-x}{x^2}.$$
40892
Is it always possible to find the limit of a function without using  or ?  For example,   $$\lim_{x\to0}\frac{\tan x-x}{x^3}$$  $$\lim_{x\to0}\frac{\sin x-x}{x^3}$$  $$\lim_{x\to0}\frac{\ln(1+x)-x}{x^2}$$  $$\lim_{x\to0}\frac{e^x-x-1}{x^2}$$  $$\lim_{x\to0}\frac{\sin^{-1}x-x}{x^3}$$  $$\lim_{x\to0}\frac{\tan^{-1}x-x}{x^3}$$
38454
I am trying to evaluate the following limit without L'Hopital's:   $$\lim_{x\rightarrow 0} \frac{\sin (6x)}{\sin(2x)}$$  I know I have to use the fact that $\frac{\sin x}{x} = 1$ but I don't know how to get the limit from the above to $\frac{\sin x}{x}$ or even a portion of it to that. I know how to evaluate limits like the following   $$\lim_{x \rightarrow 0}\frac{\sin 3x}{x} = 3$$  if that's any help.  Any hints would be appreciated  Thanks!
59163
I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use L'Hospital's rule.)  $$\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$$
12735
$$\lim_{x\to 0} \frac{\ln(1+x)}x$$  The process I  want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule)   $$\lim_{h \to 0} \frac{f(x+h) - f(x)}h$$  $$\lim_{h \to 0} \frac{\frac{\ln (1+x+h)}{x+h} - \frac{\ln(1+x)}x}h$$  $$\lim_{h \to 0} \frac{x\ln(1+x+h) - (x+h)\ln (1+x)}{hx(x+h))}$$  At this point I get confused because I know the answer is $1$, but I am not getting this answer through simplification of my formula.
52366
How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$  without using L'Hospital or Taylor series?   thanks :)
85831
By L'Hôpital's rule, it gets $$\lim_{x\to 0}-\frac{\left(e^{(x+1)^{1/x}} (x+1)^{1/x}-e (x+1)^{e/x}\right) ((x+1) \log (x+1)-x)}{2x^3(1+x)},$$which becomes more complicated.  What could I do then?
19643
The question is rather:      Prove the following:   $$  \lim_{x \to \infty} \frac{x ^ {\sqrt x} }{ 2 ^ x} = 0. $$   I was thinking of using the Squeeze Theorem (might not be the right way to go), but finding an upper-bound function proved to be quite tricky.
106621
I am trying to calculate the following limit without L'Hôpital's rule:     $$\lim_{n \to \infty} \dfrac{n}{\ln(n)}$$   I tried every trick I know but nothing works. You don't have to prove it by definition.
90878
Find the limit: $$\lim_{x\to0}{\frac{x^2\cdot\sin\frac{1}{x}}{\sin x}}$$ After treating it with l'Hopital rule, we get: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}-\cos\frac{1}{x}}{\cos x}}$$ Now, the numerator of fraction doesn't have a limit, so I can't use l'Hopital rule again. What to do? I can split it into two fractions, but I'm not sure how would it help: $$\lim_{x\to0}{\frac{2x\cdot\sin \frac{1}{x}}{\cos x}} - \lim_{x\to0}{\frac{\cos\frac{1}{x}}{\cos x}}$$
69608
I know with l'Hopital's rule it becomes $-\sin(x)$ which has the limit $0$. However, I have been wondering how to evaluate this limit without l'Hopital's rule.
71280
I was given a task to find  $$\lim_{x\to0}\frac{x-\sin{x}}{x^3}$$ at my school today. I thought it was an easy problem and started differentiating denominator and numerator to calculate the limit but the teacher then said we aren't allowed to use  L'Hopital's rule, but to "play around" with known limits and limit definition. I got stuck here since I can't really think of a way to do this, and according to my teacher, there are at least 4 ways. A subtle hint would be enough. Thanks.
66968
I need to evaluate the limit without using l'Hopital's rule.  $$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$
21939
Without using L'Hopital's rule find: $$\lim_{x\rightarrow 0}{\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}}$$ I found the first derivative because I planned to use Taylor series:  \begin{align} \frac{d}{dx}{\left(\frac{\ln{(e+x)}-e^x}{\cos^2{x} -e^x}\right)} &amp;=\frac{\left(\frac{1}{x+e}-e^x\right)\left(\cos^2{x} -e^x\right)-\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)} {(\cos^2{x} -e^x)^2}\\ &amp;=\frac{\frac{1}{x+e}-e^x}{\cos^2{x}-e^x}-\frac{\left(\ln(x+e)-e^x\right)\left(\sin{2x}-e^x\right)}{(\cos^2{x} -e^x)^2} \end{align}  However, it seems I haven't gone so far. Should I start from the beginning and try a different method?  Source in Croatian: 2.kolokvij, matematička analiza
45937
$$\lim_{x\to\infty}\left(\sin{\frac 1x}+\cos{\frac 1x}\right)^x$$  It is about the $(\to1)^{(\to\infty)}$ situation. Can we find its limit using the formula $\lim_{x\to\infty}(1+\frac 1x)^x=e$? If yes, then how?
78897
$\lim_{n \to \infty} n[(1+\frac{1}{n})^n - e]$  I let, $x = \frac{1}{n}$, then as   $\lim_{x \to 0} \frac{1}{x}[(1+x)^\frac{1}{x} - e] = \infty$  L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{1} = \frac{1}{0}(1+0)^{\frac{1}{0}-1} = \infty$  Again, if we apply L'hopital's: $\lim_{x \to 0} \frac{\frac{1}{x}(1+x)^{\frac{1}{x}-1}}{x+(1+x)}$. This is also going to $\infty$ as $x \to 0$. But, I know the answer is $\frac{-e}{2}$, and I am not even close. Can anyone please find me the mistakes here. Oh, I am supposed to use L'Hopital's rule.
91429
I have a midterm coming up and on the past exams the hard question(s) usually involve some form of $\lim_{x\to0^{+}} x \ln x$. However, we're not allowed to use l'Hopital's rule, on this year's exam anyways.  So how can I evaluate said limit without l'Hopital's rule? I got somewhere with another approach, don't know if it's useful:   $\lim_{x\to0^{+}} x \ln x = \lim_{x\to0^{+}} x^2 \ln (x^2) = L$ $= (\lim_{x\to0^{+}} 2x)(\lim_{x\to0^{+}} x \ln x)$ $= 0 * L$   Then I just need to prove that L is finite/exists (which means it must be 0)
121707
How do I prove that this limit does not exist?   $$\lim_{x\rightarrow 0} \frac{e^{1/x} - 1}{e^{1/x} + 1} $$  My attempt:  When you approach from from left towards zero , say i take -0.00000000000001 . i substitute in expression i get (-1) . But if i take 0.000000000001 and substitue i get (=1) (by applying L'Hop) . But if i dont do this and apply L'Hop straighaway i get 1 .