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This is a question from maharastra state board's commercial maths paper 2017. I have no idea how to solve this question.

If $f(x)=\frac{15^x - 3^x - 5^x + 1}{x\tan(x)}$, $x\ne0$ is continous at $x=0$, then find $f(0)$

Another query:- if $x\ne0$ then how can it be continous and how does $f(0)$ exist???

AdLibitum
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Dhruv Raghunath
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  • With regard to your "Another query", the problem could have been worded better, but the intended question was this: If f is defined for all x (including x = 0), and if, for nonzero x, the value of f(x) is given by the specified formula, what must be the value of f(0) given that f is continuous at x = 0? – quasi Mar 10 '17 at 08:54
  • Well...I think you should evaluate $\lim_{x \to 0+}f(x)$ and $\lim_{x \to 0-}f(x)$...Double application of L' Hospital's Rule gives answer to be $\ln(3)\ln(5)$ – LM2357 Mar 10 '17 at 08:54

3 Answers3

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We must have $\lim_{x\to0}f(x)=f(0)$ to be countinous, but $$\lim_{x\to0}\frac{15^x - 3^x - 5^x + 1}{x \tan(x)}=\lim_{x\to0}\dfrac{\frac{5^x-1}{x}\frac{3^x-1}{x}}{\frac{\tan x}{x}}=\ln5.\ln3$$

Nosrati
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You have to compute the limit $$\lim_{x \to 0} \frac{15^x-5^x-3^x+1}{x \tan x}$$ and this will be your $f(0)$. Now, your function can be written as $$\frac{5^x-1}{x} \cdot \frac{3^x-1}{x} \cdot \frac{x}{\sin x} \cdot \cos x$$ where each factor has finite limit. Hence your function tends to $$\log 5 \cdot \log 3 \cdot 1 \cdot 1 = \log 5 \log 3$$

This means that $$f(x) = \begin{cases} \frac{15^x-5^x-3^x+1}{x \tan x} &, &x \neq 0 \\ \log 5 \log 3 & , & x=0 \end{cases} $$ is a continuous function.

Crostul
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$$\lim_{x\to 0}f(x) = \lim_{x\to 0} \frac{15^x - 3^x - 5^x + 1}{x \tan(x)} = \lim_{x\to 0} \frac{(3^x-1)\cdot (5^x-1)}{x\cdot tan x}\cdot \frac{x}{x} = \lim_{x\to 0} \frac{(3^x-1)\cdot (5^x-1)}{x^2} = ln3\cdot ln5$$

sgrmshrsm7
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