What is wrong with this solution to the limits question?

Question

Evaluate $\lim\limits_{x\to\infty}x-x^2\ln\bigg(1+\dfrac1{x}\bigg)$.

My solution:

\begin{align} \lim\limits_{x\to\infty}x-x^2\ln\bigg(1+\frac1{x}\bigg)&=\lim\limits_{x\to\infty}x-\frac{x\ln(1+\frac1{x})}{\frac{1}x}\\ &=\lim\limits_{x\to\infty}x-(x\cdot 1)\\ &=\lim\limits_{x\to\infty}x-x\\ &=0 \end{align}

I got $0$ as answer, but the correct answer is $\frac12$.

I solved it using another method, but I just need to know why this won't work. Any ideas?

Sorry, it is actually 1/2, I just wrote in a hurry. So why was I getting 0. — Shweta Banerjee, Sep 21 '19 at 06:24
Do you know multiplication law of limit? i.e., when can you distribute limit over multiplication? — IamKnull, Sep 21 '19 at 06:27
Yes, and I was trying to apply that here, but I still don't see how this won't work. — Shweta Banerjee, Sep 21 '19 at 06:28
No, the multiplication law is applicable only when both limits exist and are finite. Check why this is not the case with your solution. — IamKnull, Sep 21 '19 at 06:30
Try to find this limit $\lim_{x\to\infty}x\cdot\frac{1}{x}$ by distributing the limit. — IamKnull, Sep 21 '19 at 06:31
Well my point is you can't write $\lim_{x\to\infty}x\cdot\frac{1}{x}$ as $\lim_{x\to\infty}x\cdot\lim_{x\to\infty}\frac{1}{x}$ Check this propetry and codition for clear understanding — IamKnull, Sep 21 '19 at 06:34
And please do not use images in lieu of text: they can cause problems with some displays, they are generally not searchable within the site. Check How do I type math in my question/answer/comment? — IamKnull, Sep 21 '19 at 06:36
@Shweta, set $1/x=h$ see third problem in https://math.stackexchange.com/questions/387333/are-all-limits-solvable-without-lh%c3%b4pital-rule-or-series-expansion — lab bhattacharjee, Sep 21 '19 at 06:45
Replacing a sub-expression by its limit is allowed in specific cases but not in general. It is best to understand when such replacements are / are not allowed. See details in https://math.stackexchange.com/a/1783818/72031 — Paramanand Singh, Sep 21 '19 at 17:20

score 3 · Answer 1 · answered Sep 21 '19 at 06:57

When you went from $$ \lim_{x \rightarrow \infty} x - \left( x \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}} \right) $$ to $$ \lim_{x \rightarrow \infty} x - \left( x \cdot 1 \right) $$ you must have passed through an expression where "$ \lim_{x \rightarrow \infty} \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}}$" appeared. (Otherwise, how did you replace that subexpression with its limit?)

Pretty much the only way to do that is along \begin{align*} \lim_{x \rightarrow \infty} \; & \left( x - \left( x \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}} \right) \right) \\ \qquad &\overset{?}{=} \left( \lim_{x \rightarrow \infty} x \right) - \left( \lim_{x \rightarrow \infty} x \right) \left( \lim_{x \rightarrow \infty} \frac{\ln \left( 1+\frac{1}{x} \right)}{\frac{1}{x}} \right) \text{,} \end{align*} which is hopeless, because two of those limits do not exist. So the equality can (and as you have found, does) fail.

Always remember, the various versions of $$ \lim_{\dots} {\dots} = \left( \lim_{\dots} \dots \right) [\text{operation}] \left( \lim_{\dots} \dots \right) $$ all require that the limits on the right exist for equality to be guaranteed.

By this logic, we can't take the limit inside the subtraction in the first place right, as that will encounter one term which leads to infinity and the other which is indeterminate? — Shweta Banerjee, Sep 21 '19 at 07:25
Agreed. (The indeterminate is the source of the "surprise" in this problem.) — Eric Towers, Sep 21 '19 at 07:31

score 2 · Answer 2 · answered Sep 21 '19 at 10:22

$$ \begin{align} \lim_{x\to\infty}\left(x-x^2\log\left(1+\frac1{x}\right)\right) &=\lim_{x\to\infty}\left(x-\frac{x\log\left(1+\frac1x\right)}{\frac1x}\right)\tag1\\ &=\lim_{x\to\infty}\left(x-x\lim_{x\to\infty}\frac{\log\left(1+\frac1x\right)}{\frac1x}\right)\tag2\\ &=\lim_{x\to\infty}x\lim_{x\to\infty}\left(1-\frac{\log\left(1+\frac1x\right)}{\frac1x}\right)\tag3\\ \end{align} $$ Step $(1)$ is fine. Step $(2)$ is where the argument in the question goes wrong. It is not legal to simply apply a limit to a piece inside another limit. Step $(3)$ is fine and shows that step $(2)$ introduced a limit of the form $\infty\cdot0$.

score 1 · Answer 3 · answered Sep 21 '19 at 06:24

1

Write your term in the form $$\frac{\frac{1}{x}-\ln\left(1+\frac{1}{x}\right)}{\frac{1}{x^2}}$$ and use the rules of L'Hospital.

answered Sep 21 '19 at 06:24

Dr. Sonnhard Graubner

95,283

Thank you so much, I did end up using LH rule. But I don't understand why the solution I provided here won't work. Do you think you can decode that? – Shweta Banerjee Sep 21 '19 at 06:27
Since one term is of the form $\infty \times 0$, this is undetermined. – Dr. Sonnhard Graubner Sep 21 '19 at 06:32
Thank you. I understood. – Shweta Banerjee Sep 21 '19 at 06:38

Z Ahmed · Answer 4 · 2019-09-21T10:14:23.083

1

$$L=\lim_{x\rightarrow \infty} x\left(1-x\ln(1+1/x) \right) \Rightarrow x (1-x(\frac{1}{x}-\frac{1}{2} \frac{1}{x^2} + \frac{1}{3} \frac{1}{x^3}..))=\lim_{x \rightarrow \infty}(\frac{1}{2}-\frac{1}{3} \frac{1}{x})=\frac{1}{2}.$$

edited Sep 21 '19 at 10:14

answered Sep 21 '19 at 08:39

Z Ahmed

43,235

Where did the $1/n$ come from and go to? Perhaps you meant $1/x$. – robjohn Sep 21 '19 at 08:51
OH! yes it is $1/x$. – Z Ahmed Sep 21 '19 at 10:14

What is wrong with this solution to the limits question?

4 Answers4