2
$\begingroup$

I know the answer for the following limit calculation, but I do not know how it is evaluated step by step. Could you please give me a hand on this one?

$${\lim\limits_{x \to 0}{{e^{\tan^2(x)} - \cos(x)} \over \tan^2(x)}} = {3 \over 2}$$

Thank you.

$\endgroup$
5
$\begingroup$

Hint: $\dfrac{e^{\tan^2x}-\cos x}{\tan^2x}=\dfrac{e^{\tan^2x}-1}{\tan^2x}+\dfrac{1-\cos x}{\tan^2x}$, and use the fact that $\dfrac{e^x-1}{x} \to 1$ as $x \to 0$.

$\endgroup$
  • $\begingroup$ Oh i see and $ {{ 1 - cos(x) } \over { x^2 }} = {1 \over 2 } $ , thank you. $\endgroup$ – Attila Horváth Jan 2 '16 at 21:12
  • $\begingroup$ The big O notation is good but you only use it as your last resource. $\endgroup$ – DeepSea Jan 2 '16 at 21:13
4
$\begingroup$

HINT:

$$\begin{align} \cos x&=1-\frac12x^2+O(x^4)\\\\ \tan x&=x+O(x^3)\\\\ e^{\tan ^2x}&=1+\tan^2 x+O(\tan^4x)\\\\ &=1+x^2+O(x^3) \end{align}$$

$\endgroup$
  • $\begingroup$ Thank you for the hint, I am going to look into the the big O notation since I have never used it before. $\endgroup$ – Attila Horváth Jan 2 '16 at 21:25
  • $\begingroup$ You're welcome. My pleasure. Happy New Year. - Mark $\endgroup$ – Mark Viola Jan 2 '16 at 21:41
1
$\begingroup$

You could use expansion of the functions as Dr.MV explained. You can also use the L'hospital rule which says if the Numerator and Denominator both tend to zero when we put the limit, keep differentiating both numerator and denominator separately until you reach a stage when either the numerator or denominator do not tend to zero.

i.e.

$${\lim\limits_{x \to 0}{e^{tan^2(x)}−cos(x) \over {tan^2(x)}} ={2e^{tan^2(x)}sec^2(x)tan(x) +sin(x) \over {2tan(x) sec^2(x)}}}$$

Differentiate again and put x=0.

This is easier than remembering the expansions.

$\endgroup$
  • $\begingroup$ Thank you for the help on L'hospital's rule, I was trying to find a simpler, shorter solution to the problem even though this is really helpful too. $\endgroup$ – Attila Horváth Jan 2 '16 at 21:30

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.