Limit question regarding logarthmic function.

Question

$\displaystyle{\lim_{x \to 0}}f(x)$

where $f(x)$=$\frac{\ln(1+x^2+x)+\ln(1+x^2-x)}{secx-cosx}$

What I found was the two ways

Way 1

$L$=$\frac{\frac{\ln(1+x^2+x)(x+x^2)}{(x+x^2)}+\frac{\ln(1+x^2-x)(x^2-x)}{(x^2-x)}}{secx-cosx}$

$L$=$\frac{2x^2}{secx-cosx}$$=$$2$

Way2-

$\frac{\ln(1+x^2+x)+\ln(1+x^2-x)}{secx-cosx}$

$L$=$\frac{\ln((1+x^2+x^4)}{secx-cosx}$

$L$=$\frac{(x^2+x^4)}{secx-cosx}$=1

Which method is correct and why?

Answer 1

You could use the notable limits :

$\lim\limits_{t\to0}\dfrac{\ln(1+t)}t=1\;\,,\quad$ $\lim\limits_{x\to0}\dfrac{\sin x}x=1\;.\quad\color{blue}{(*)}$

Now, we will calculate your limit by using $\,(*)\,$:

$\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x\right)+\ln\left(1+x^2-x\right)}{\sec x-\cos x}=$

$=\lim\limits_{x\to0}\dfrac{\ln\left[\left(1+x^2+x\right)\left(1+x^2-x\right)\right]}{\sec x-\cos x}=$

$=\lim\limits_{x\to0}\dfrac{\ln\big[\left(1+x^2\right)^2-x^2\big]}{\sec x-\cos x}=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{\sec x-\cos x}=$

$=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{\frac1{\cos x}-\cos x}=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)\!\cdot\!\cos x}{1-\cos^2\!x}=$

$=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)\!\cdot\!\cos x}{\sin^2\!x}=$

$=\lim\limits_{x\to0}\left[\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}\!\cdot\!\dfrac{x^2\left(1+x^2\right)\cos x}{\sin^2\!x}\right]=$

$=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}\!\cdot\!\lim\limits_{x\to0}\left(\!\dfrac x{\sin x}\!\right)^2\!\!\cdot\lim\limits_{x\to0}\big[\!\left(1+x^2\right)\cos x\big]=$

$=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}\!\cdot\!1^2\!\cdot\!1=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}\,.$

By letting $\;t=x^2+x^4\,,\;$ it follows that $\;t\to0\;$ as $\;x\to0\;$ and consequently we get that

$\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}=\lim\limits_{t\to0}\dfrac{\ln(1+t)}t=1\,.$

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Addendum :

It is wrong to calculate your limit in the following way :

$\lim\limits_{x\to0}\dfrac{\left(x^2+x\right)\frac{\ln\left(1+x^2+x\right)}{x^2+x}+\left(x^2-x\right)\frac{\ln\left(1+x^2-x\right)}{x^2-x}}{\sec x-\cos x}=$

$=\lim\limits_{x\to0}\dfrac{\left(x^2+x\right)\!\cdot\!1+\left(x^2-x\right)\!\cdot\!1}{\sec x-\cos x}=$

$=\lim\limits_{x\to0}\dfrac{\left(x^2+x\right)+\left(x^2-x\right)}{\sec x-\cos x}\,.$

It is wrong because you have calculated the limit of just a part of the function without using any theorem about the calculation of limits.

I have calculated the limit of a part of the function too when I wrote that

$\lim\limits_{x\to0}\left[\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}\!\cdot\!\dfrac{x^2\left(1+x^2\right)\cos x}{\sin^2\!x}\right]=$

$=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}\!\cdot\!\lim\limits_{x\to0}\left(\!\dfrac x{\sin x}\!\right)^2\!\!\cdot\lim\limits_{x\to0}\big[\!\left(1+x^2\right)\cos x\big]=$

$=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}\!\cdot\!1^2\!\cdot\!1=\lim\limits_{x\to0}\dfrac{\ln\left(1+x^2+x^4\right)}{x^2+x^4}\,,$

but what I did is correct because I applied the theorem that states that the limit of a product is equal to the product of the limits (unless there is the indeterminate form $0\!\cdot\!\infty$).

But which theorem have you applied in your method that you called “way 1” ?

Answer 2

By Taylor's formula, $$\log(1+x^2+x)+\log(1+x^2-x)=\log(1+x^2+x^4)=x^2+O(x^4)$$ and $$\sec x=1+\frac{x^2}{2}+O(x^4) \\ \cos x=1-\frac{x^2}{2}+O(x^4)$$ all as $x\to 0$, hence, $$\frac{\log(1+x^2+x)+\log(1+x^2-x)}{\sec x-\cos x}=\frac{x^2+O(x^4)}{x^2+O(x^4)}$$ as $x\to0$. The limit is $1$.