3
$\begingroup$

I managed to solve part of this limit but can't get the final step right. Here's the limit: $$ \lim_{x\to0} { \frac { \left( 1+\sin{x}+\sin^2{x} \right) ^{1/x} - \left( 1+\sin{x} \right) ^{1/x} } { x } } $$ Let's begin then.
Using $f(x) = e^{\log{f(x)}} = \exp{\left[\log{f(x)}\right]}$ it gets to: $$ \lim_{x\to0} { \frac { \exp \left[ \frac { \log{\left(1+\sin{x}+\sin^2{x}\right)} } { x } \right] - \exp \left[ \frac { \log{\left(1+\sin{x}\right)} } { x } \right] } { x } } $$ Multiplying and dividing by the right terms: $$ \lim_{x\to0} { \frac { \exp \left[ \frac { \log{\left(1+\sin{x}+\sin^2{x}\right)} } { \sin{x}+\sin^2{x} } \frac { \sin{x}+\sin^2{x} } { x } \right] - \exp \left[ \frac { \log{\left(1+\sin{x}\right)} } { \sin{x} } \frac { \sin{x} } { x } \right] } { x } } $$ Grouping the $\sin{x}$ in the first exponential function: $$ \lim_{x\to0} { \frac { \exp \left[ \frac { \log{\left(1+\sin{x}+\sin^2{x}\right)} } { \sin{x}+\sin^2{x} } \frac { \sin{x} } { x } \left(1+\sin{x}\right) \right] - \exp \left[ \frac { \log{\left(1+\sin{x}\right)} } { \sin{x} } \frac { \sin{x} } { x } \right] } { x } } $$ Applying the know limits, specifically:
$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}+\sin^2{x}\right)}}{\sin{x}+\sin^2{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\log{\left(1+\sin{x}\right)}}{\sin{x}}=1}$$$$\lim \limits_{x\to0}{\frac{\sin{x}}{x}=1}$$$$\lim \limits_{x\to0}{\left(1+\sin{x}\right)}=1$$

We should end up with: $$ \lim_{x\to0} { \frac { \exp \left[ \frac { \log{\left(1+\sin{x}+\sin^2{x}\right)} } { \sin{x}+\sin^2{x} } \frac { \sin{x} } { x } \left(1+\sin{x}\right) \right] - \exp \left[ \frac { \log{\left(1+\sin{x}\right)} } { \sin{x} } \frac { \sin{x} } { x } \right] } { x } } = \frac { \exp{\left[1(1)(1)\right]} - \exp{\left[1(1)\right]} } { 0 } = \frac{e-e}{0} = \frac{0}{0} $$ Undetermined form, yay. Any hint is really appreciated, and I'm really really sorry for all the spaghetti rendering, it's hard to look at.

$\endgroup$
6
$\begingroup$

Evaluate $$L=\lim_{x\to0} {\frac{\left(1+\sin{x}+\sin^2{x}\right)^{1/x}-\left(1+\sin{x}\right)^{1/x}}{x}}$$

If there are more terms to apply the limit to, you should separate the terms of which limits are apparent to reduce the probability of such failure.


\begin{align} L&=\lim_{x\to0} {\frac{\left(1+\sin{x}+\sin^2{x}\right)^{1/x}-\left(1+\sin{x}\right)^{1/x}}{x}}\\ &=\lim_{x\to0} \left(1+\sin{x}\right)^{1/x}{\frac{\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)^{1/x}-1}{x}}\tag{1}\\ &=e\lim_{x\to0} {\frac{\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)^{1/x}-1}{x}}\tag{2}\\ &=e\lim_{x\to0} {\frac{\exp\left[\frac 1x\ln\left(1+\frac{\sin^2{x}}{1+\sin{x}}\right)\right]-1}{x}}\tag{3}\\ &=e\lim_{x\to0} {\frac{\exp\left\{\displaystyle\frac 1x\left[\sum_{i=1}^{+\infty}(-1)^{i+1}\frac1i\left(\frac{\sin^2{x}}{1+\sin{x}}\right)^i\right]\right\}-1}{x}}\tag{4}\\ &=e\lim_{x\to0} {\frac{{\displaystyle\sum_{j=1}^{+\infty}\frac{\left\{\displaystyle\frac 1x\left[\sum_{i=1}^{+\infty}(-1)^{i+1}\frac1i\left(\frac{\sin^2{x}}{1+\sin{x}}\right)^i\right]\right\}^j}{j!}}}{x}}\tag{5}\\ &=e\lim_{x\to0} {\frac{\frac 1x\frac{\sin^2{x}}{1+\sin{x}}}{x}}=e\tag{6}\\ \end{align} where in step $(6)$, I used the term $i=j=1$ only, because

  • $\displaystyle\lim_{x\to 0}(1+\sin x)=1$.
  • $\displaystyle\lim_{x\to 0}\frac{\sin^a x}{x^b}=1\Leftrightarrow a=b$, where $a,b\in\mathbb R$.
  • $a\rightarrow 2ij=b\rightarrow j+1\Rightarrow j(2i-1)=1$, where $i,\ j\geq 1$

You did something like $\displaystyle\lim_{x\to 0}\frac{e-e}{0}$ in the place of my step $(4)$ because of overwhelmingly many terms on both sides of the minus sign.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Sorry for asking but just to make sure: in $(4)$ you approximated the logarithm? $\endgroup$ – Gaetano Esposito Jul 7 at 15:02
  • $\begingroup$ As pointed by @GaetanoEsposito you should add more steps and not resort to such approximations. $\endgroup$ – Paramanand Singh Jul 8 at 0:00
  • $\begingroup$ @GaetanoEsposito I have tried to avoid all ambiguity now. Have a look at the edited! $\endgroup$ – Sameer Baheti Jul 8 at 6:15
  • $\begingroup$ @ParamanandSingh Thanks for the feedback. I edited the solution to be on the safer side. $\endgroup$ – Sameer Baheti Jul 8 at 6:16
  • $\begingroup$ Thanks for considering my feedback, but you took the hard route. You can do it more simply by putting $t=\sin^2x/(1+\sin x)$ and $u=(1/x) \log(1+t)$ so that both $u, t$ tend to $0$ and then $$\frac{e^u-1}{x}=\frac{e^u-1}{u}\frac{u}{x}$$ Since first factor tends to $1$ you just deal with $u/x=(1/x^2)\log(1+t)$ or $$\frac{t} {x^2}\cdot\frac{\log(1+t)}{t}$$ and both factors tend to $1$. $\endgroup$ – Paramanand Singh Jul 8 at 6:25
2
$\begingroup$

Hint: Write the function of interest as $$ (1+\sin x)^\frac{1}{x}\,\frac{e^{\frac{1}{x}\log(1+\frac{\sin^2 x}{1+\sin x})}-1}{x} $$ and remember that $$ \lim_{x\to0}\frac{\sin x}{x}=1\,,\qquad \lim_{x\to0}\frac{\log(1+x)}{x}=1\,,\qquad \lim_{x\to0}\frac{e^{x}-1}{x}=1\,. $$ You should find that the limit is $e$.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Your second last equality is not justified: You cannot take limit of enumerator and denominator separately as the limit of the denominator is (trivially) equal to 0. Try the appropriate version of l'Hospitals rules (which helps, because enumerator and denominator (as you have formally shown) converge to $0$.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

You should note that the expression in numerator is of the form $A-B$ with both $A, B$ tending to $e$. We can write $$A-B=B\cdot\frac{\exp(\log A-\log B) - 1}{\log A-\log B} \cdot(\log A - \log B) $$ Since the middle factor tends to $1$ the desired limit is equal to the limit of $$e\cdot\frac{\log A-\log B} {x} =e\cdot\frac{\log(1+\sin x+\sin^2x)-\log(1+\sin x)} {\sin^2x}\cdot\frac{\sin^2x}{x^2}$$ The last factor tends to $1$ and putting $t=\sin x$ we see that desired limit equals the limit of $$e\cdot\frac{\log(1+t+t^2)-\log(1+t)}{t^2}=e\cdot\frac{\log(1+(t^2/(1+t)))}{t^2/(1+t)}\cdot\frac{1}{1+t}$$ and clearly the above tends to $e\cdot 1\cdot 1=e$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.