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Application of L'Hopital: TIFR GS 2018 Part A, Problem 2

Understand the problem

True or false: $\lim _{x \rightarrow 0} \frac{\sin x}{\log (1+\tan x)}=1$

Start with hints

Do you really need a hint? Try it first!

"Hint 1"
  • Observe that the following limit is of the form \(\frac 00\).
  • Do you remember that we always solve the limits of the form \(\frac 00\) and \(\frac{\infty}{\infty}\) by L’Hospital’s Rule.
  • Hint 2
    • Consider \(f(x)=sinx\) and \(g(x)=log(1+tanx)\)
    • Compute \(f ‘(x)=cosx\) and \(g ‘(x) = sec^2(x)/(1+tanx)\)
    • Compute the limit of \(f ‘(x)/g ‘(x)\)
    Hint 3
    • Observe that the \(lim f ‘(x)=1\) and \(lim g ‘(x) =1 \).
    • Hence the value of the Limit is \(1\).
    • The statement is therefore TRUE.

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