$$\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$$

I used L'Hospital's rule, but after second application it is still not possible to determine the limit. When applying Taylor series, I get wrong result ($\frac{-1}{6}$). What method to use?

Result should be $\frac{1}{4}$

  • 1
    $\begingroup$ multiply nominator and denominator by $$\sqrt{1+\tan(x)}+\sqrt{1+x}$$ $\endgroup$ Aug 17, 2015 at 20:13
  • 4
    $\begingroup$ The result should actually be $0$ not $1/4$. Are you sure you wrote down the problem correctly? $\endgroup$
    – JimmyK4542
    Aug 17, 2015 at 20:14
  • $\begingroup$ @JimmyK4542 Maybe it is a mistake in my book's solution. $\endgroup$
    – user300045
    Aug 17, 2015 at 20:15

5 Answers 5


L'Hospital's rule is not the α and ω of limits computation. First remove the square roots in the numerator: $$\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=\frac{\tan x-x}{(\sqrt{1+\tan x}+\sqrt{1+x})\sin^2x}$$ Now use equivalents:

  • $\tan x-x=x+\dfrac{x^3}3+o(x^3)-x$, hence $\;\tan x-x\sim_0 \dfrac{x^3}3$
  • $\sqrt{1+\tan x}+\sqrt{1+x}\xrightarrow[x\to 0]{}2$
  • $\sin x\sim_0 x$

$$\text{So we have:}\hspace{8em}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}\sim_0\frac{\dfrac{x^3}3}{2x^2}=\frac x6\to 0.\hspace{8em}$$


Using Taylor series we have

$$\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=\frac{\tan x-x}{\sin^2x(\sqrt{1+\tan x}+\sqrt{1+x})}\sim_0\frac{\frac13x^3}{2x^2}=\frac1{6}x$$ so the desired limit is $0$.


$$\lim_{x \to 0} \dfrac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=L$$

Using L'hopital:

$$\lim_{x \to 0} \dfrac{\dfrac{\sec^2x}{2\sqrt{1+\tan x}}-\dfrac{1}{2\sqrt{1+x}}}{2\sin x \cos x}=L$$

Reordering the denominator: $$\lim_{x \to 0} \dfrac{\dfrac{\sec^2x}{2\sqrt{1+\tan x}}-\dfrac{1}{2\sqrt{1+x}}}{\sin 2x }=L$$

Using L'hopital Again

$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x }=L$$

You cannot use l'hopital again because $\lim_{x\to 0} \cos 2x = 1 > 0$

so replacing:

$$\lim_{x \to 0} \dfrac{\dfrac{-2 \sec^2 x\tan x \sqrt{1+\tan x}-\frac{\sec^4 x}{2\sqrt{1+\tan x}}}{2(1+\tan x)}+\dfrac{1}{4\sqrt{1+x}^3}}{2\cos 2x } = \dfrac{\frac{0-\frac{1}{2}}{2}+\frac{1}{4}}{2\cdot 1} = 0$$


If you know that $$\lim_{x \to 0}\frac{\tan x - x}{x^{2}} = 0\tag{1}$$ then it is easy to give a simple evaluation for the limit in question \begin{align} L &= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{\sin^{2}x}\notag\\ &= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{x^{2}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\ &= \lim_{x \to 0}\frac{\sqrt{1 + \tan x} - \sqrt{1 + x}}{x^{2}}\notag\\ &= \lim_{x \to 0}\frac{\tan x - x}{x^{2}\{\sqrt{1 + \tan x} + \sqrt{1 + x}\}}\notag\\ &= \frac{1}{2}\lim_{x \to 0}\frac{\tan x - x}{x^{2}}\notag\\ &= 0\notag \end{align}


This can be (almost) completely solved using algebraic manipulation:

$$\frac{\sqrt{\tan x+1}-\sqrt{x+1}}{\sin^2(x)} = \frac{\tan x -x}{\sin^2 x\cdot(\sqrt{\tan x+1} +\sqrt{x+1})}$$

Now, since: $$\lim_{x\to 0}\frac{1}{\sqrt{\tan x+1} +\sqrt{x+1}}=1$$

Our problem reduces to: $$\lim_{x\to 0}\frac{\tan x -x}{\sin^2 x}= \lim_{x\to0}\frac{1}{\sin x}\bigg(\frac{\tan x}{\sin x}-\frac{x}{\sin x}\bigg)=$$

$$\lim_{x\to0}\frac{1}{\sin x}\cdot\lim_{x\to0}\bigg(\frac{\tan x}{\sin x}-\frac{x}{\sin x}\bigg)$$

Also, $$\lim_{x\to0}\frac{\sin x}{x} = \lim_{x\to 0}\frac{\tan x}{\sin x} = 1$$

Then: $$\lim_{x\to0}\frac{1}{\sin x}\cdot\lim_{x\to0}\bigg(\frac{\tan x}{\sin x}-\frac{x}{\sin x}\bigg) = \lim_{x\to 0}\frac{1}{\sin x}\cdot 0 = 0$$

I am aware of the issue I have in my solution - most Calculus instructors I know would accept this work for this question.

  • $\begingroup$ At the university I attend, you would lose at least one point for the last step. You can't separate the limits s. t. one of them exists, while the other is $\infty$. Ignorance of the instructors isn't a justification. Answer by Paramamand Singh is already complete and you unnecessarily activated an old question. $\endgroup$
    – Invisible
    Sep 28, 2020 at 9:29

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