How to Evaluate $\lim_{x \to 0} \frac{\cos(x) - \ln(1+x) - \sqrt{1+2x}}{x^2}$ Without L'Hôpital or Taylor Expansion?

Question

it's my first Analysis class in college. I'm trying to evaluate the following limit:

$$\lim_{x \to 0} \frac{\cos(x) - \ln(1+x) - \sqrt{1+2x}}{x^2}$$

but I would like to avoid using L'Hôpital's Rule or Taylor series expansions.

Is there a more "direct" or classical method to approach this limit? Perhaps by using other mathematical tools? Any insights or detailed steps would be greatly appreciated.

L'Hôpital and Taylor are the classical approaches in this case, I don't see how you would solve it in another way — Vincent Batens, Dec 07 '24 at 15:38
Using Taylor expansion is the best strategy here, why do you want to avoid it? — Sine of the Time, Dec 07 '24 at 15:38
You can get rid of $\cos x$ by subtracting and adding $1$ in the numerator and by applying $$\cos x-1=-{\sin^2x\over 1+\cos x}$$ But with remaining part of the numerator $1-\ln(1+x)-\sqrt{1+2x}$ I do not see an elementary way for solving the limit. — Ryszard Szwarc, Dec 07 '24 at 15:44
Isn't "analyzing the dominant terms" the same as using Taylor expansion? — Sine of the Time, Dec 07 '24 at 15:45
It's what the professor told us. He prohibited us from using those techniques. — CHOSM, Dec 07 '24 at 16:05

lab bhattacharjee · Accepted Answer · 2024-12-07T18:57:55.483

6

Hint:

$$\dfrac{\cos x-\ln(1+x)-\sqrt{1+2x}}{x^2}$$

$$=-\dfrac{1-\cos x}{x^2}-\dfrac{\ln(1+x)-x}{x^2}+\dfrac{1-x-\sqrt{1+2x}}{x^2}$$

The first one is easy and should be left as an exercise.

For the second one, use Are all limits solvable without L'Hôpital Rule or Series Expansion .

For the third, rationalize the numerator.

edited Dec 07 '24 at 18:57

answered Dec 07 '24 at 16:22

lab bhattacharjee

279,016

It should be $-ln(1+x)$ not $ln(1+x)$ – CHOSM Dec 07 '24 at 16:28
1

@CHOSM, Thanks for your observation – lab bhattacharjee Dec 07 '24 at 18:14
1

and $1+x-\sqrt{1+2x}$ should be $1-x-\sqrt{1+2x}$ now – CHOSM Dec 07 '24 at 18:55
1

@CHOSM, Thanks again, for a finite result, we should have $$\sqrt{1-2x}$$ – lab bhattacharjee Dec 07 '24 at 18:58

score 1 · Answer 2 · answered Dec 07 '24 at 17:36

You don’t need to think to far, look at the questions more carefully and you’ll see that you don’t need L'Hôpital's Rule or Taylor series expansion to solve this problem $$\lim_{x \to 0} \frac{\cos(x) - \ln(1+x) - \sqrt{1+2x}}{x^2}$$ The sum (or minus) at the numerator doesn’t matter because the three functions are expected to relate to each other and cancel out the factor $\frac{1}{x}$ or powers of it

$$\lim_{x \to 0} \{ \frac{\cos(x) }{x^2}- \frac{\ln(1+x) }{x^2}-\frac{\sqrt{1+2x} }{x^2} \}$$ By observation none of the three parts contains $x^2$ in their numerator and obviously all three diverges, even if you’re not sure of the $\cos()$ part and $\ln()$ part see that $\sqrt()$ part $$\lim_{x \to 0} \sqrt{ \frac{1}{x^4}+\frac{2}{x^3} }$$ It contains the forbidden $\frac{1}{x}$ and because the limit is held by a sum, if one part diverges so does the rest

Tell your Professor the limit doesn’t exist as it diverges to infinity

score 0 · Answer 3 · answered Dec 07 '24 at 17:14

Thank you so much guys for your prompt answers.

Thanks to the answer of lab bhattacharjee , I have come to the solution: To evaluate:

$$ \lim_{x \to 0} \frac{\cos x - \ln(1+x) - \sqrt{1+2x}}{x^2}, $$

without using L’Hôpital's Rule or Taylor expansion, we can proceed by decomposing the expression as follows:

$$ \frac{\cos x - \ln(1+x) - \sqrt{1+2x}}{x^2} = \frac{1 - \cos x}{x^2} + \frac{-\ln(1+x) + x}{x^2} + \frac{1 - x - \sqrt{1+2x}}{x^2}. $$

Now, let’s handle each term separately.

1. $\frac{1 - \cos x}{x^2}$

This is a well-known trigonometric limit: $$ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = -\frac{1}{2}. $$

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

This limit has been tackled here and it is equal to $\frac{1}{2}.$

3. $\frac{1 - x - \sqrt{1+2x}}{x^2}$

For this term, we are going to multiply the numerator and denominator by the conjugate $(1 - x + \sqrt{1+2x})$: $$ \frac{1 - x - \sqrt{1+2x}}{x^2} = \frac{(1 - x - \sqrt{1+2x})(1 - x + \sqrt{1+2x})}{x^2(1 - x + \sqrt{1+2x})} = \frac{x^2 - 4x}{x^2(1 - x + \sqrt{1+2x})}. = \frac{x-4}{x(1 - x + \sqrt{1+2x})}. $$

As $x \to 0$, the denominator $x(1 - x + \sqrt{1+2x}) \to 0$, while the numerator $(x-4) \to -4$. Thus, this term diverges to $\infty$.

Now combine all terms: $$ \lim_{x \to 0} \frac{\cos x - \ln(1+x) - \sqrt{1+2x}}{x^2} = \infty. $$

How to Evaluate $\lim_{x \to 0} \frac{\cos(x) - \ln(1+x) - \sqrt{1+2x}}{x^2}$ Without L'Hôpital or Taylor Expansion?

3 Answers3

1. $\frac{1 - \cos x}{x^2}$

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

3. $\frac{1 - x - \sqrt{1+2x}}{x^2}$

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