Home › Forums › Math questions › Calculus questions

Limits

Posted December 15th, 2009 by alpha
in

$ \mathop{\lim}\limits_{x \to 0}\frac{10^x-2^x-5^x+1}{xtanx} $

‹ Trigonometric limits Limits ›

Comments

stuck at this point-kindly check.

On December 15th, 2009 alpha says:

Adding and subtracting 1 in the numerator and grouping in the following way-
$ =>\mathop{\lim}\limits_{x \to 0}\frac{(10^x -1)-(2^x-1)-(5^x-1)}{xtanx} $
$ =>\mathop{\lim}\limits_{x \to 0}\frac{(10^x -1)-(2^x-1)-(5^x-1)}{x} X \frac{1}{tanx} $
There after I can use this rule(for only the expression lying left to multiplication sign):
$ \mathop{\lim}\limits_{x \to0}\frac{a^x-1}{x}=ln a=log_e a $
How to deal with $ \frac{1}{tanx} $ ?
?
?

-------------------x----------------------------
``Old theorems never die; they turn into definitions.''
----E. Hewitt

''ALPHA" =)

answer

On December 15th, 2009 Structure says:

$ \mathop{\lim}\limits_{x \to 0}\frac{10^x-2^x-5^x+1}{xtanx}=\lim_{x\to 0}\frac{(5^x-1)(2^x-1)}{x^2}\frac{x}{\tan x}=\ln5*\ln 2 $

Back to top