Question

If the function $f$ defined as $f\left(x\right)=\frac{1}{x}-\frac{k-1}{e^{2x}-1}$, $x\ne 0$, is continuous at $x=0$, then the ordered pair $(k,f(0\left)\right)$ is equal to?

• A. $(3,1)$
• B. $(3,2)$
• C. $\left(\frac{1}{3},2\right)$
• D. $(2,1)$

Answer 1

The correct option is A $(3,1)$

If the function is continuous at $x=0$, $\underset{x\to 0}{lim}f\left(x\right)$ exists and is equal to $f\left(0\right)$

$\underset{x\to 0}{lim}f\left(x\right)$

$=\underset{x\to 0}{lim}\frac{1}{x}-\frac{k-1}{e^{2x}-1}$

$=\underset{x\to 0}{lim}\frac{e^{2x}-1-kx+x}{\left(x\right)(e^{2x}-1)}$

Using Taylor's expansion for $e^{2x}$, we get

$=\underset{x\to 0}{lim}\frac{(1+2x+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^3}{3!}+...)-1-kx+x}{\left(x\right)\left(\right(1+2x+\frac{(2x{)}^2}{2!}+\frac{(2x{)}^3}{3!}+...)-1)}$

$=\underset{x\to 0}{lim}\frac{(3-k)x+\frac{4x^2}{2!}+\frac{8x^3}{3!}+...}{(2x^2+\frac{4x^3}{2!}+\frac{8x^4}{3!}+...)}$

For the limit to exist, power of $x$in the numerator should be greator than or equal to the power of $x$ in the denominator.

Therefore, coefficient of $x$ in numerator is equal to zero

$⟹3-k=0$

$⟹k=3$

So the limit reduces to

$\underset{x\to 0}{lim}\frac{\left(x^2\right)(\frac{4}{2!}+\frac{8x}{3!}+...)}{\left(x^2\right)(2+\frac{4x}{2!}+\frac{8x2}{3!}+...)}$

$=\underset{x\to 0}{lim}\frac{\frac{4}{2!}+\frac{8x}{3!}+...}{2+\frac{4x}{2!}+\frac{8x2}{3!}+...}$

$=1$

Hence, $f\left(0\right)=1$

Therefore, answer is option (A).