Question

Question on the application of l'Hopital's rule is to follow. Please wait :)

Answer 1

\[\lim_{x \rightarrow 0}\frac{x-\sin x}{x-\tan x}\]

Answer 2

So, take the derivative of the numerator and the denominator, yeah?

Answer 3

Where are you getting stuck?

Answer 4

The result of taking the derivative of the numerator and the denominator is that the ratio is indeterminate for x=0.

Answer 5

Yeah, you have to use L'Hopitals twice in this case.

Answer 6

The result of a second differentiation is:\[\frac{\sin x}{-\sec ^{2}x \tan x}\]
This ratio is also indeterminate for x=0.

Answer 7

\[\frac{x-\sin(x)}{x-\tan(x)}\implies \frac{\cos(x)-1}{\tan^2(x)} \implies =-\frac12\cos^3(x)\]
\[\lim -\frac12\cos^3(x)=-\frac12\lim\cos^3(x)=-\frac12*1=-\frac12\]

Answer 8

\[\lim_{x \rightarrow 0}\frac{x-\sin x}{x-\tan x}\\
\lim_{x \rightarrow 0}\frac{1-\cos x}{1-\sec^2 x}\\
\lim_{x \rightarrow 0}\frac{\sin x}{-2\sec^2 x\tan x}=\lim_{x \rightarrow 0}\frac{\sin x \ \cos^2 x}{-2\tan x}=\lim_{x \rightarrow 0}-\frac{1}{2}\cos^3x=-\frac{1}{2}\\
\]

Answer 9

When I differentiate numerator and denominator of :\[\frac{\sin x}{-2\sec ^{2}x \tan x}\]the result is as follows:\[\frac{\cos x}{-2(\sec ^{4}x+2\sec ^{2}x \tan ^{2}x)}\]For x=0 this gives the value:\[-\frac{1}{2}\]

Answer 10

just differentiating once is enough...just cancel out the sinx from numerator with the tanx ,using tanx=sinx.secx....

Answer 11

It has been necessary to differentiate three times to get to f'''(x) before reaching a result for the limit following l'Hopital's rule.

Answer 12

It wasn't necessary for you to differentiate three times. While differentiating three times does work, you can simplify the second derivative to obtain the same result.

Answer 13

Thank you for that :)