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Ok so I have issues with this specific 'type' of limits. : $\\$

$$\lim_{x\to1}(4^x-3^x)^{\frac{1}{1-x}}$$

and $\\$

$$\lim_{x\to-1}\biggr(-4\cdot\arctan(x)\cdot{\frac{1}{\pi}}\biggr)^{\frac{1}{x+1}}$$ $\\$

It seems like they are quite similar, but I'm not sure what to do. I've tried taking $\ln$ of the limit to simplify it but have reached nothing. I would appreciate any hint whatsoever.

However, I am not allowed to use L'hospital rule or integrals to solve these.

Thanks in advance!

Collapse
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    Are you allowed to use the definition of the derivative, i.e., $$f^{\prime}(x) = \lim_{t \to x}\dfrac{f(t)-f(x)}{t-x}\text{?}$$ – Clarinetist Jan 22 '17 at 15:25
  • Nope, no derivatives at all, the problem specifically says so, and we haven't reached derivations at all yet at uni.. – Collapse Jan 22 '17 at 15:26
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    Could you give us an idea of what you've been allowed to use to solve limits? Given the information above, I'm not sure if you're looking for anything more than "here's a graph of the equation and here's what it looks like around that $x$-value, so this is the limit." – Clarinetist Jan 22 '17 at 15:28
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    Well we used different kinds of identites, like e.g. $\lim_{x\to0}(\frac{e^x-1}{x})=1$ and identities for $sin(x) / x$ and stuff like that. Maybe taking $ln$ of that limit, and then later returning the value back. I need to make an algebraic proof of how I reached the specific limit – Collapse Jan 22 '17 at 15:30
  • Thank you for the information you've provided. I'm not sure how I would approach this problem given what you've stated, but perhaps someone else might have a better idea. – Clarinetist Jan 22 '17 at 15:32
  • No problems, thanks for trying, I hope so too. :/ – Collapse Jan 22 '17 at 15:33
  • Theres a problem with the second limit the base is negative so taking it to a power is not well defined. – Rene Schipperus Jan 22 '17 at 15:52
  • @ReneSchipperus pardon I wrote it wrong, I've edited it now! – Collapse Jan 22 '17 at 16:00
  • Ok well the same technique as I used will work, you might want to note that $\arctan(y+1)-\arctan(1)=\arctan(\frac{y}{2+y})$. – Rene Schipperus Jan 22 '17 at 16:16
  • I see, thanks!! – Collapse Jan 22 '17 at 16:42

2 Answers2

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Change the variable to $x=y+1$ and you have now $y\to 0$. Next take the log so you have

$$-\frac{\ln(4^{y+1}-3^{y+1})}{y}$$

Now use that $$\frac{\ln(4^{y+1}-3^{y+1})}{4^{y+1}-3^{y+1}-1}\to 1$$ as a special instance of $\lim\limits_{t\to 0}\frac{\ln 1+t}{t}=1$.

So the problem reduces to $$-\frac{4^{y+1}-3^{y+1}-1}{y}$$ which we write as $$\frac{3^{y+1}-3}{y}-\frac{4^{y+1}-4}{y}=3\frac{3^{y}-1}{y}-4\frac{4^{y}-1}{y}$$

Finally we use the fact that $$\frac{a^t-1}{t}\to \ln a$$ to see that your limit is $$3\ln 3-4\ln 4$$

But dont forget to undo the log so the real limit is $$\frac{3^3}{4^4}$$

Rene Schipperus
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  • Thanks a lot for your answer! I just don't quite understand how you reached the form of your third row, I'm aware of the special instance identity you wrote but how do I apply it exactly? – Collapse Jan 22 '17 at 15:43
  • Ok look at the second displayed equation, the denominator is $t=4^{y+1}-3^{y+1}-1$ which I use in the identity, its important that this expression goes to zero. To go from the first displayed equation to the third I multiplied by the reciprocal of the second expression. – Rene Schipperus Jan 22 '17 at 15:48
  • Wow can't believe I didn't see this, thanks a lot, this has been really helpful! – Collapse Jan 22 '17 at 15:50
  • Glad to hear it. – Rene Schipperus Jan 22 '17 at 15:51
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I thought it might be instructive to present an approach that relies only on a set of elementary inequalities. To that end, we begin with a primer.

PRIMER:

In THIS ANSWER, I used only the limit definition of the exponential function and Bernoulli's Inequality to show that the exponential function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{1+x\le e^x\le \frac{1}{1-x}} \tag 1$$

for $x<1$, and the logarithm function satisfies the inequalities

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1} \tag2$$

for $x>0$

We note that without loss of generality, we assume that the limit is taken from the left-hand side (i.e., $x<1$). Proceeding, let $f(x)=(4^x-3^x)^{1/(1-x)}$. Then, using $t=e^{\log(t)}$ we can write $f(x)$ as

$$\begin{align} f(x)=e^{\frac{\log(4^x-3^x)}{1-x}} \tag 3 \end{align}$$

Letting $g(x)=\frac{\log(4^x-3^x)}{1-x}$ and exploiting $(1)$ and $(2)$, the upper bound for $g(x)$ is

$$\begin{align} g(x)&\le \frac{4^x-3^x-1}{1-x}\\\\ &=\frac{4e^{(x-1)\log(4)}-3e^{(x-1)\log(3)}-1}{1-x}\\\\ &\le \frac{\frac{4}{1-(x-1)\log(4)}-3(1+(x-1)\log(3))-1}{1-x}\\\\ &=\frac{3\log(3)-4\log(4)-3\log(3)\log(4)(x-1)}{(1-(x-1)\log(4))}\\\\ &\to \log\left(\frac{3^3}{4^4}\right)\,\,\,\text{as}\,\,x\to 1 \tag 4 \end{align}$$

and the lower bound for $g(x)$ is

$$\begin{align} g(x)&\ge \frac{4^x-3^x-1}{(1-x)(4^x-3^x)}\\\\ &=\frac{4e^{(x-1)\log(4)}-3e^{(x-1)\log(3)}-1}{(1-x)(4^x-3^x)}\\\\ &\ge \frac{4(1+(x-1)\log(4))-\frac{3}{1-(x-1)\log(3)}-1}{(1-x)(4^x-3^x)}\\\\ &=\frac{3\log(3)-4\log(4)+4\log(3)\log(4)(x-1)}{4^x-3^x}\\\\ &\to \log\left(\frac{3^3}{4^4}\right)\,\,\,\text{as}\,\,x\to 1 \tag 5 \end{align}$$

Applying the squeeze theorem to $(4)$ and $(5)$ shows that $\lim_{x\to 1}g(x)=\log\left(\frac{3^3}{4^4}\right)$ whereupon using $(3)$ and exploiting the continuity of the exponential function yields the coveted limit

$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 1}(4^x-3^x)^{1/(1-x)}}=\frac{3^3}{4^4}$$

Note that the only tools used were the inequalities in $(1)$ and $(2)$ along with the continuity of the exponential function and the squeeze theorem!

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Mark Viola
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