The questions is: Evaluate $$\lim_{x\to 0} \frac{a^x -1}{x}$$ without applying L'Hopital's Rule.  Does this question fundamentally same as asking if the $\lim_{x\to 0} \frac{a^x -1}{x}$ exists? rather than straightway asking to find the limit. That means are questions (1) proving if the limit of a function exists and (2) asking what is the limit of that function, essentially same question?
15636
Taking as our definition of exponentiation repeated multiplication (extended to real exponents by continuity), can we show that the limit   $$\lim_{h\to 0}\dfrac{a^h-1}{h}$$  exists, without l'Hôpital, $e$, or even natural logarithm? Sure, l'Hôpital will work, but that's circular if we're developing calculus of transcendentals from first principles. There is  to this question already by user Neal, but he uses the exponential function with base $e$ (it's been answered many times: see also , , and ).   But using the special properties of $e$ strikes me as circular too; not literally logically circular, in the sense that we are invoking results we're trying to prove, since there are definitions of $e^x$ which make it trivial to verify the derivative. But perhaps pedagogically circular for a complete novice, the special properties of $e$ appear unmotivated because they cannot be justified without reference to the very derivative we are trying to compute (or else a detour through logarithms, but let's not).  Can we find nice squeeze theorem bounds like Neal has, but for the function $a^x$ instead of $e^x$, with the additional handicap that we can't just write $a^x=e^{x\log a}$? I thought to substitute a series expansion for $\log a$, but didn't come up with any bounds that were nicely polynomial in both $x$ and $a$.  I wonder whether the geometric proof of $\lim (\sin x)/x$ (see for example, robjohn's answer ) could be adapted.  Obviously without a reference to natural logarithm, we cannot compute the value of the limit. But I just want to show it exists (via squeeze theorem or monotone convergence). Once we know this limit exists, we can show it behaves like a logarithm, whence there is a unique base for which the limit is 1, which we call $e$. The rest of the development of calculus of exponentials and logs follows easily. This seems like the approach that would appear the most accessible yet motivated to a novice calculus student.   An analogous limit to $\lim\dfrac{a^h-1}{h}$ for understanding to differentiating exponential functions, are the limits $\lim\limits_{n\to\infty} (1+\frac{1}{n})^n$ and $\lim\limits_{n\to\infty} (1+\frac{1}{n})^{n+1}$ for differentiating the logarithm, if you prefer to start with that as your primitive concept. Both limits are shown to exist using  (see WimC's answer  for the first limit, and see David Mitra's answer  for the second limit). I tried without success to use Bernoulli's inequality to show my sequence was monotone. This limit can also be analyzed using  as seen in user94270's answer to . So that inequality may help here.  I would also accept an explanation of why the limit cannot be computed without transcendental techniques, or an opinion why this is not a pedagogically sound approach to introducing the calculus of exponentials and logarithms.  Edit:  has a nice solution by Paramanand Singh to a closely related problem.
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I was wondering what is the value of $\lim\limits_{x\to 0+} x^{x}$.  Assuming that the limit exists, I could show using the usual logarithm techniques that the limit is $1$. However, I am not able to show that the limit exists. Could some one help on that?  EDIT: It would be so nice if we could do it without l'Hopital...
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I am studying limits at infinity, and I have a doubt about evaluating them. From what I know, limits only exist if both sides of the limit exist and are equal. For example, take a look at the following limit:  $$\lim_{ x\to \infty}\frac{x}{e^{-x}}$$  As $x \to +\infty$ the function goes to $+\infty$. However, as $x \to -\infty$ the function goes to $0$.   Does this mean that the limit does not exist or I only have to evaluate the limit as x goes to $+\infty$?
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Today in my analysis class, we were preparing for the final and this question came up:     Evaluate $$\lim_{x\to0}\frac{e^{-1/x^2}}{x}$$   We tried taking the $\log$, using L'Hopitals and some other tricks but couldn't figure it out.   I thought maybe viewing this limits as a sequence in the following way might help;  $$\lim_{x\rightarrow0}\frac{e^{-1/x^2}}{x} = \lim_{n\rightarrow \infty}\frac{e^{-n^2}}{1/n} = \lim_{n\rightarrow \infty}\frac{n}{e^{n^2}}$$  But from them I'm not sure.   Thank you.
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I'm trying to solve this limit     $$\lim_{x\to 0} \frac{\sin x-x}{x^3}$$   Solving using L'hopital rule, we have:  $$\lim_{x\to 0} \frac{\sin x-x}{x^3}= \lim_{x\to 0} \frac{\cos x-1}{3x^2}=\lim_{x\to 0} \frac{-\sin x}{6x}=\lim_{x\to 0} \frac{-\cos x}{6}=-\frac{1}{6}.$$  Am I right?  I'm trying to solve this using change of variables, I need help.  Thanks  EDIT  I didn't understand the answer and the commentaries, I'm looking for an answer using change of variables.
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I need to evaluate the limit without using l'Hopital's rule.  $$\lim_{x \to 0}{\frac{\sqrt{1+x\sin{x}}-\sqrt{\cos{2x}}}{\tan^{2}{\frac{x}{2}}}}$$
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Does the limit $\lim _{x\to 0} \frac 1x \int_0^x \left|\cos \frac 1t \right| dt$ exists ? If it does then what is the value ?   I don't think even L'Hospital's rule can be applied . Please help . Thanks in advance
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How do I evaluate this limit: $$\lim\limits_{x\to \infty}\frac{\sin x}{x}$$  Is it $0$? If yes, how so?
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Consider the Limit  $\frac{x+\sin x}{x-\sin x}$  as $x  \rightarrow \infty$. Clearly this limit exists and is equal to $1$. Also notice that the function is of the form  $\frac{\infty}{\infty}$ and thus, L'Hopitals rule can be applied here. But on applying L'Hopitals rule, I am getting  Limit  $\frac{1+\cos x}{1-\cos x}$ as $x  \rightarrow \infty$, which is obviously non-converging. Where did I go wrong?
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 :)
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What is the limit of $0/x$ as x goes to $0$, without using L'Hopital's rule? Clearly it should be $0$, but I'm not sure how it is any different from something like the divergence of $1/r^2$ which yields a dirac delta.
20178
How do you evaluate the limit of  $$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3}$$  by first principle? I was thinking of this way to evaluate but it seems to evaluate to $\frac {0} {0}$.  $$\lim_{x \to 9} \frac {x-9} {\sqrt{x} - 3} = \frac{\lim_{x \to 9} (x-9) }{\lim_{x \to 9}  {\sqrt{x} - 3}}$$  I mentioned that the limits do not exist because the function is not continuous at $ x=9$. Any idea on how to solve this question?
74526
How to find the limit of: $$ \lim_{x\to \infty }x\left(\sqrt[x]{a}-1\right)$$ Without using L'hôspital rule.  I've tried to bound the term and use the squeeze theorem but I couldn't find the right upper bound. I've also tried to convert $a^\frac{1}{x}$ to $e^{\frac{1}{x}\ln{a}}$ but it didn't help me.  Whats the right way to evaluate that limit?
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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 .
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So I ran into this problem:  $\lim _{x\to \pi }(\frac{\cos\left(\frac{x}{2}\right)}{\pi ^2-x^2})$  We haven't studied L'Hôpital's rule in my class yet, we basically need to transform the expression using trigonometric identities and use known limits like $\lim_{x \to 0} \frac{sinx}{x} = 1$ to calculate in case we get an indefinite form.  I couldn't find anything I could do to calculate this limit :(  Some help?  Thanks.
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I read in Stewart "single variable calculus" page 83 that the limit $$\lim_{x\to 0}{1/x^2}$$ does not exist. How precise is this statement knowing that this limit is $\infty$?. I thought saying the limit does not exist is not true where limits are $\infty$. But it is said when a function does not have a limit at all like  $$\lim_{x\to \infty}{\cos x}$$.
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I have heard people say that you can't (or shouldn't) use the L'Hopital's rule to calculate $\lim\limits_{x\to 0}\frac{\sin x}x=\lim\limits_{x\to 0}\cos x=1$, because the result $\frac d{dx}\sin x=\cos x$ is derived by using that limit.  But is that opinion justified? Why should I be vary of applying L'Hopital's rule to that limit?  I don't see any problem with it. The sine function fulfills the conditions of the L'Hopital's  rule.  Also, it is a fact that the derivative of sine is cosine, no matter how we proved it. Certainly there is a way to prove $\frac d{dx}\sin x=\cos x$ without using the said limit (if someone knows how, they can post it) so we don't even have any circular logic. Even if there isn't,  $\frac d{dx}\sin x=\cos x$ was proven sometime without referencing the L'Hopital's rule so we know it is true. Why wouldn't we then freely apply the L'Hopital's rule to $\frac {\sin x}x$?  PS I'm not saying that this is the best method to derive the limit or anything, but that I don't understand why it is so frowned upon and often considered invalid.