Evaluating the Limit of Cosine Function Using L'Hospital's Rule - Explained

[navneet9431]

<Moderator's note: Moved from a technical forum and thus no template.>
$$\lim_{x \to 0} \cos(\pi/2\cos(x))/x^2$$
I tried to evaluate the limit this way,
$$\lim_{x \to 0} \cos(\pi/2\cdot1)/x^2$$ since $$\cos0=1$$
$$\lim_{x \to 0} \cos(\pi/2\cdot1)/x^2=\lim_{x \to 0} 0/x^2$$
Now apply L'Hospital's Rule twice,
$$\lim_{x \to 0} 0/2(x)=\lim_{x \to 0} 0/2=0$$
So,this way the answer is zero.
Can you please explain where am I doing wrong?
I will be thankful for help!

 

[Phylosopher]

You cannot just take the limit for one part and leave the rest. Either you take the limit for the whole expression or not.

This would be long, but use the L'Hopital rule twice on the original expression to remove x^2 from the denominator

 

[ehild]

You cannot just take the limit for one part and leave the rest. Either you take the limit for the whole expression or not.

This would be long, but use the L'Hopital rule twice on the original expression to remove x^2 from the denominator

It s enough to apply L'Hopital rule once. You will get a product, and arrange the factors, so they have known limits. (One factor will be sin(x)/x.)

 

[Delta2]

The product rule for limits $\lim{f(x)g(x)}=\lim{f(x)}\lim{g(x)}$ holds only if both the limits of $f(x)$ and $g(x)$ are finite or $+-\infty$. If one limit is 0 and the other is infinite, as happens in this case where $\lim_{x\to 0}\cos(\frac{\pi}{2}\cos x)=0$ and $\lim_{x\to 0}{\frac{1}{x^2}}=\infty$, then the product rule cannot be applied.

 

[Ray Vickson]

<Moderator's note: Moved from a technical forum and thus no template.>
$$\lim_{x \to 0} \cos(\pi/2\cos(x))/x^2$$
I tried to evaluate the limit this way,
$$\lim_{x \to 0} \cos(\pi/2\cdot1)/x^2$$ since $$\cos0=1$$
$$\lim_{x \to 0} \cos(\pi/2\cdot1)/x^2=\lim_{x \to 0} 0/x^2$$
Now apply L'Hospital's Rule twice,
$$\lim_{x \to 0} 0/2(x)=\lim_{x \to 0} 0/2=0$$
So,this way the answer is zero.
Can you please explain where am I doing wrong?
I will be thankful for help!

If you write $\cos x = 1- A$ for small $|x|$, your given fraction is
$$\text{fraction} = \frac{\sin(\pi/2 \: A)}{x^2}.$$ Now look at how $A = 1 - \cos x$ behaves for small $|x|$, or use l'Hospital's rule.