2

Once again, I am not interested in the answer. But rather, where is/are my mistake(s)? Perhaps the solution route is hopeless:

Question is: evaluate $\lim_{x \to 4} \frac{x^4 -4^x}{x-4}$.

My workings are:

Let $y=x-4$. Then when $x \to 4$, we have that $y \to 0$. Thus:

$$\lim_{y \to 0} \frac{(y+4)^4 - 4^{y+4}}{y} = \\ = \lim_{y \to 0}\frac{(y+4)^4}{y} - \lim_{y \to 0} \frac{4^{(y+4)}}{y} $$

And this step is not allowed from the get go, as I am deducting infinities, which is indeterminate. What I should have done though:

$$4^4 \lim_{y \to 0} \frac{(1+y/4)^4-1+(4^y-1)}{y} = \\ 4^4 \lim_{y \to 0} \left( \frac{(1+y/4)^4-1}{\frac{y}{4}4} - \frac{4^y-1}{y} \right) = \\ =4^4\left(\frac{1}{4} \cdot 4 - \ln 4 \right) = 256(1-\ln 4)$$

enter image description here

Naz
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  • why downvote?.. – Naz Oct 15 '18 at 20:47
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    I didn't downvote, but if you expect people to check your work, it would be courteous to take the time to type it up and make it much easier to read. – operatorerror Oct 15 '18 at 20:49
  • noted. Thanks.. – Naz Oct 15 '18 at 20:52
  • @isquared-KeepitReal Read here – Gibbs Oct 15 '18 at 20:52
  • I will rewrite it tomorrow. Sorry for trouble. I usually do type it. – Naz Oct 15 '18 at 20:52
  • Since answer was already given, I'll just mention that you can use l'opital's rule – Makina Oct 15 '18 at 20:52
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    It's simply the derivative of $f(x)=x^4-4^x$ at $x=4$; since $f'(x)=4x^3-4^x\log4$, we have $f'(4)=256-256\log4=256(1-\log 4)$. – egreg Oct 15 '18 at 20:54
  • @egreg That's of course the best way to solve it! – user Oct 15 '18 at 21:16
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    @gimusi I always tell my students that knowing derivatives is knowing many limits. I find a bit silly forcing people to use low level methods when better ones are available. – egreg Oct 15 '18 at 21:17
  • @egreg Yes I agree with you, once we know derivatives we can use effectively that tool to solve limits such that. Of course in my opinion that mehod is preferable to l'Hopital or Taylor in that case. For me an exception are standard limits which aredefined before derivative concept and are also a powerful tool to solve limits. – user Oct 15 '18 at 21:29
  • I found another mistake. We have: $\displaystyle{\lim_{x \to 0} \frac{(1+x)^p-1}{x}} = p$. I am hopeless. I keep making these silly mistakes. I should retire from maths. – Naz Oct 15 '18 at 21:32
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    @isquared-KeepitReal I didn't check for all the mistakes since the first one is sufficient to obtain an uncorrect result! Do not discourage for that. I can suggest to solve many limits as you can to enforce your skills on that. – user Oct 15 '18 at 21:34

3 Answers3

4

The step where you rewrite the limit as the difference of two other limits ((i) and (ii)) is not legitimate. You can only equate a limit to a sum or difference of two limits if both those limits converge.

David K
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  • If we use addition on $R\cup {\pm \infty}$ we can also extend the addition rules to the case $+\infty+\infty=+\infty$ and $-\infty-\infty=-\infty$ but maybe it is not the best rigorous way to proceed. – user Oct 15 '18 at 21:07
  • In this case we have $+\infty - \infty.$ But I did neglect to say how in other cases one might use a sum or difference to prove a sum diverges in a particular direction. – David K Oct 15 '18 at 21:16
  • That's just a detail, your answer is very clear to me. – user Oct 15 '18 at 21:20
3

The following step is not allowed

$$\lim_{x \to 4} \frac{x^4-4^x}{x-4}=\lim_{y \to 0} \frac{(y+4)^4-4^{y+4}}{y}\color{red}{=\lim_{y \to 0} \frac{(y+4)^4}{y}-\lim_{y \to 0} \frac{4^{y+4}}{y}}$$

Refer also to the related

user
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  • It is not exactly "not allowed" but it creates and indeterminate and is not helpful. – Doug M Oct 15 '18 at 21:06
  • @DougM It is not allowed...in that case of course. The link I gave discuss the topic in great detail. – user Oct 15 '18 at 21:08
  • @DougM When I claim that it is not allowed I mean that $$\lim_{y \to 0} \frac{(y+4)^4-4^{y+4}}{y}\color{red}{\neq}\lim_{y \to 0} \frac{(y+4)^4}{y}-\lim_{y \to 0} \frac{4^{y+4}}{y}$$ – user Oct 15 '18 at 21:15
1

You can break up this limit under certain circumstances.

$\lim_\limits{x\to a} (f(x) + g(x)) = \lim_\limits{x\to a} f(x) + \lim_\limits{x\to a}g(x)$

You can do it if $\lim_\limits{x\to a} f(x), \lim_\limits{x\to a}g(x)$ both exist and are finite.

But if $\lim_\limits{x\to a} f(x) = \infty$ and $\lim_\limits{x\to a}g(x) = -\infty,$ then you have just given yourself and indeterminate form.

You could do this:

$\lim_\limits{y\to 0} \frac {(y+4)^4 - 4^{y+4}}{y} = \lim_\limits{y\to 0} \frac {y^4}{y}+ \lim_\limits{y\to 0} \frac {4y^3}{y}+\lim_\limits{y\to 0} \frac {6y^2}{y} +\lim_\limits{y\to 0} \frac {4y}{y} + \lim_\limits{y\to 0} \frac {4^4(1 - 4^y)}{y}$

Because each of those limits are defined and finite.

You could also use L'Hopital's rule.

Doug M
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  • so far I've done this: $4^4\lim_{y \to 0} \frac{(1+y/4)^4-4^y}{y}$. And stuck again. I'm trying to solve it without the derivatives. – Naz Oct 15 '18 at 21:10
  • lacking for $4$ in the denominator for the "first part" and "minus one" in the numerator for the "second part". But cannot make those two things "friends". – Naz Oct 15 '18 at 21:11
  • I get $\displaystyle{4^4 \lim_{y \to 0} \left( \frac{(1+y/4)^4}{4}\cdot \frac{4}{y}-\frac{(4^y-1)+1}{y} \right)}$ – Naz Oct 15 '18 at 21:12
  • I do need to rewrite them there: $\displaystyle{4^4 \lim_{y \to 0} \left( \frac{(1+y/4)^4-1}{\frac{y}{4}4} - \lim_{y \to 0} \frac{4^y-1}{y} \right)}$ – Naz Oct 15 '18 at 21:19
  • The term on the left... just multiply it out. The term on the right evaluates to $\ln 4$ – Doug M Oct 15 '18 at 21:24