Question

If \[f(x)=\left\{ \begin{align}
  & \dfrac{{{\cos }^{2}}x-{{\sin }^{2}}-1}{\sqrt{{{x}^{2}}+1}-1},x\ne 0 \\
 & k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x=0 \\
\end{align} \right.\] is continuous at $x=0$, find $k$.

Answer 1

Hint: A function $f(x)$ is continuous at $x=c$($c$ being any arbitrary constant), only if the left hand limit of $f(x)$at $x=c$(x tending to c from the negative side), the right hand limit of $f(x)$ at $x=c$(x tending to c from the positive side) and $f(c)$ are all equal.

Complete step-by-step answer:
Mathematically, if $f(x)$ is continuous at $x=c$,
$\displaystyle \lim_{x \to {{c}^{-}}}f(x)=f(c)=\displaystyle \lim_{x \to {{c}^{+}}}f(x)$
In the question when x tends to zero, finding the value of limit from either of the two sides will suffice, as both of them are equal. That is, $\displaystyle \lim_{x \to {{0}^{-}}}f(x)=\displaystyle \lim_{x \to {{0}^{+}}}f(x)=\displaystyle \lim_{x \to 0}f(x)$

We have,
\[f(x)=\left\{ \begin{align}
  & \dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x-1}{\sqrt{{{x}^{2}}+1}-1},x\ne 0 \\
 & k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x=0 \\
\end{align} \right.\]is continuous at $x=0$.
We know that, $\displaystyle \lim_{x \to {{c}^{-}}}f(x)=f(c)=\displaystyle \lim_{x \to {{c}^{+}}}f(x)$
At $x=0$,
$f(x)=f(0)=k................(1)$
In this case, the left hand limit is equal to the right hand limit.
Hence, on taking the limits for x tending to 0, we get,
$\displaystyle \lim_{x \to 0}f(x)=\displaystyle \lim_{x \to 0}\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x-1}{\sqrt{{{x}^{2}}+1}-1}$
We know the trigonometric identity$\cos 2x={{\cos }^{2}}x-{{\sin }^{2}}x$. Using this trigonometric identity the above equation becomes
$\displaystyle \lim_{x \to 0}f(x)=\displaystyle \lim_{x \to 0}\dfrac{\cos 2x-1}{\sqrt{{{x}^{2}}+1}-1}$
We also know the trigonometric identity$\cos 2x=1-2{{\sin }^{2}}x$. Using this trigonometric identity we obtain
$\displaystyle \lim_{x \to 0}f(x)=\displaystyle \lim_{x \to 0}\dfrac{-2{{\sin }^{2}}x}{\sqrt{{{x}^{2}}+1}-1}$
On rationalizing,
$\displaystyle \lim_{x \to 0}f(x)=\displaystyle \lim_{x \to 0}\dfrac{-2{{\sin }^{2}}x}{\sqrt{{{x}^{2}}+1}-1}\left( \dfrac{\sqrt{{{x}^{2}}+1}+1}{\sqrt{{{x}^{2}}+1}+1} \right)$
Using the algebraic identity ${{a}^{2}}-{{b}^{2}}=(a+b)(a-b)$ in the denominator,
$\displaystyle \lim_{x \to 0}f(x)=\displaystyle \lim_{x \to 0}\dfrac{-2{{\sin }^{2}}x\left( \sqrt{{{x}^{2}}+1}+1 \right)}{{{x}^{2}}+1-1}$
$=-2\displaystyle \lim_{x \to 0}\left( \dfrac{{{\sin }^{2}}x}{{{x}^{2}}} \right)\left( \sqrt{{{x}^{2}}+1}+1 \right)$
$=-2\displaystyle \lim_{x \to 0}{{\left( \dfrac{\sin x}{x} \right)}^{2}}\left( \sqrt{{{x}^{2}}+1}+1 \right)$
We know that, $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=1$
On substituting $x$ with zero and $\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}$ as 1 the equation becomes,
\[\displaystyle \lim_{x \to 0}f(x)=-2{{(1)}^{2}}\left( \sqrt{{{0}^{2}}+1}+1 \right)\]
\[\therefore \displaystyle \lim_{x \to 0}f(x)=-2\times 2=-4\,\,\,.................(2)\]
Since $\displaystyle \lim_{x \to {{c}^{-}}}f(x)=f(c)=\displaystyle \lim_{x \to {{c}^{+}}}f(x)$, equation (1) is equal to equation (2).
 In other words,
$\displaystyle \lim_{x \to 0}f(x)=f(0)$
$\therefore k=-4$
Hence, the value of k when \[f(x)=\left\{ \begin{align}
  & \dfrac{{{\cos }^{2}}x-{{\sin }^{2}}-1}{\sqrt{{{x}^{2}}+1}-1},x\ne 0 \\
 & k\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,,x=0 \\
\end{align} \right.\] is continuous at x is equal to -4.

Note: Alternate method:
We can solve for the limit value using L hospitals’ rule as well.
Consider the equation $\displaystyle \lim_{x \to c}\dfrac{f(x)}{g(x)}$
According to L hospitals’ rule, whenever an infinity condition occurs (a divided by zero condition, that is, when the denominator tends to zero), we have to differentiate the numerator and denominator separately and then substitute for x. In the above equation, if g(c) is equal to zero we have an infinity condition. Then, using L hospitals’ rule,
\[\displaystyle \lim_{x \to 0}\dfrac{f(x)}{g(x)}=~\displaystyle \lim_{x \to 0}\dfrac{\dfrac{d}{dx}\left( f(x) \right)}{\dfrac{d}{dx}\left( g(x) \right)}\]
If the infinity condition persists we can continue the chain of differentiation.
Mathematically,
  \[\displaystyle \lim_{x \to 0}\dfrac{f(x)}{g(x)}=~\displaystyle \lim_{x \to 0}\dfrac{\dfrac{d}{dx}\left( f(x) \right)}{\dfrac{d}{dx}\left( g(x) \right)}=~\displaystyle \lim_{x \to 0}\dfrac{\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( f(x) \right)}{\dfrac{{{d}^{2}}}{d{{x}^{2}}}\left( g(x) \right)}=\displaystyle \lim_{x \to 0}\dfrac{\dfrac{{{d}^{3}}}{d{{x}^{3}}}\left( f(x) \right)}{\dfrac{{{d}^{3}}}{d{{x}^{3}}}\left( g(x) \right)}\,...........\,\]and so on.