In this post we are going to solve INDETERMINATE form \(\displaystyle \;\; \frac{0}{0}\).
We want to solve \(\displaystyle \lim_{x \to c}\frac{f(x)}{g(x)}\).
But when you fill the limit you get
\(\displaystyle \lim_{x \to c} f(x)=0\;\)
and
\(\displaystyle\lim_{x \to c}g(x)=0\;\),
then above form will bacomes a \(\;\frac{0}{0}\;\). form.
Which is one of INDETERMINATE forms, to solve this we make use L'Hôpital's rule, How to use L'Hôpital's rule we have discussed in this post.
\(\displaystyle 1).\;\; \lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\)
Solution : Put the limit, we will get
\(\displaystyle \frac{\sin a(0)}{\operatorname{sinb}(0)}=\frac{0}{0}\)
This is a \(\displaystyle \frac{0}{0}\) form so we well use L'Hôpital's rule
\(\displaystyle \Rightarrow \lim _{x \rightarrow 0} \displaystyle \frac{\displaystyle \frac{d}{d x}(\sin a x)}{\displaystyle \frac{d}{d x}(\sin b x)} \)
\(\displaystyle \Rightarrow \lim _{x\rightarrow 0} \displaystyle \frac{a \cos a x}{b \cos b x}\)
Now put limit, we will get
\(\displaystyle \Rightarrow \frac{a \times \cos (0)}{b \times \cos (0)}=\frac{a}{b}\)
ANSWER :\(\displaystyle \;\; \frac{a}{b}\)
\(\displaystyle 2).\;\; \lim _{x \rightarrow 0} \displaystyle \displaystyle \frac{x \cos x-\log (1+x)}{x^{2}}\)
Solution : Put the limit we will get
\(\displaystyle \frac{0 \times \cos (0)-\log (1+0)}{0}=\frac{0}{0}\)
This is a \(\displaystyle \;\; \frac{0}{0}\;\;\)form we will Use L'Hôpital's rule Here
\(\displaystyle \Rightarrow \lim _{x \rightarrow 0} \frac{\displaystyle \frac{d}{d x}[x \cos x-\log (1+x)]}{\displaystyle \frac{d}{d x}\left(x^{2}\right)}\)
\(\displaystyle \lim _{x \rightarrow 0} \frac{-x \sin x+\cos x- \displaystyle \frac{1}{1+x}}{2 x} \;\;\;\;\cdots (2)\)
Put limit we will get
\(\displaystyle \Rightarrow \frac{-0 \times 0+1- \displaystyle \frac{1}{1}}{2 \times 0}=\frac{0}{0}\)
again \(\displaystyle \frac{0}{0}\) form
Again apply L'Hôpital's rule on (2) we will get
\(\displaystyle \lim _{x \rightarrow 0} \frac{-x \cos x-\sin x-\sin x+ \displaystyle \frac{1}{(1+x)^{2}}}{2}\)
Put the limits
\(\displaystyle \frac{-0 \times 1-0-0+1}{2}=\frac{1}{2}\)
ANSWER :\(\displaystyle \;\; \frac{1}{2}\)
\(\displaystyle 3).\;\; \lim _{x \rightarrow 0} \frac{x e^{x}-\log (1+x)}{x}\)
Solution : Put limit, We will get
\(\displaystyle \frac{0 \times e^{0}-\log (1)}{0}=\frac{0}{0}\)
Using L'Hospital's Rule, We will get
\(\displaystyle \lim _{x \rightarrow 0} \frac{\displaystyle \frac{d}{d x}\left[x e^{x}-\log (1+x)\right]}{\displaystyle \frac{d}{d x}(x)}\)
\(\displaystyle \lim _{x \rightarrow 0} \frac{ x e^{x}+e^{x}-\displaystyle \frac{1}{1+x}}{1}\)
Put the Limit
\(\displaystyle \frac{0+e^{0}-\frac{1}{1+0}}{1}=\frac{1-1}{1}=0\)
\(\displaystyle \Rightarrow \lim _{x \rightarrow 0}\displaystyle \frac{\cos x}{2}\)
Now put limit we get \(\;\displaystyle \frac{\cos (0)}{2} =\frac{1}{2}\)
ANSWER :\(\;\displaystyle \frac{1}{2}\)
\(\displaystyle \frac{0+e^{0}-\frac{1}{1+0}}{1}=\frac{1-1}{1}=0\)
ANSWER :\(\displaystyle\;\; 0\)
\(\displaystyle 4).\;\; \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}\)
Solution : If you put limit ,then you will see that, this is \(\displaystyle \;\frac{0}{0} \;\) form
Using L'Hôpital's rule
\(\displaystyle\Rightarrow \lim _{x \rightarrow 0} \frac{\displaystyle \frac{d}{d x}(1-\cos x)}{\displaystyle \frac{d}{d x}\left(x^{2}\right)}\)
\(\displaystyle \Rightarrow \lim _{x \rightarrow 0} \displaystyle \frac{\sin x}{2 x}\)
\(\displaystyle 4).\;\; \lim _{x \rightarrow 0} \frac{1-\cos x}{x^{2}}\)
Solution : If you put limit ,then you will see that, this is \(\displaystyle \;\frac{0}{0} \;\) form
Using L'Hôpital's rule
\(\displaystyle\Rightarrow \lim _{x \rightarrow 0} \frac{\displaystyle \frac{d}{d x}(1-\cos x)}{\displaystyle \frac{d}{d x}\left(x^{2}\right)}\)
\(\displaystyle \Rightarrow \lim _{x \rightarrow 0} \displaystyle \frac{\sin x}{2 x}\)
Now put the limit
again this is \(\displaystyle \frac{0}{0}\) form
Again apply L'Hôpital's Rule
\(\displaystyle \Rightarrow \lim _{x \rightarrow 0} \frac{\displaystyle \frac{d}{d x}(\sin x)}{\displaystyle \frac{d}{d x}(2x)} \)
again this is \(\displaystyle \frac{0}{0}\) form
Again apply L'Hôpital's Rule
\(\displaystyle \Rightarrow \lim _{x \rightarrow 0} \frac{\displaystyle \frac{d}{d x}(\sin x)}{\displaystyle \frac{d}{d x}(2x)} \)
\(\displaystyle \Rightarrow \lim _{x \rightarrow 0}\displaystyle \frac{\cos x}{2}\)
Now put limit we get \(\;\displaystyle \frac{\cos (0)}{2} =\frac{1}{2}\)
ANSWER :\(\;\displaystyle \frac{1}{2}\)
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