$\displaystyle \lim_{x\rightarrow 0}\frac{sin^2x-x^2cosx}{x^3tanx}$ $\displaystyle \lim_{x\rightarrow \infty}\frac{2^{x+1}+3^{x+1}}{2^x+3^x}$ In the second one how can i change the $\displaystyle x\rightarrow \infty$
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Originally Posted by srirahulan $\displaystyle \lim_{x\rightarrow \infty}\frac{2^{x+1}+3^{x+1}}{2^x+3^x}$ In the second one how can i change the $\displaystyle x\rightarrow \infty$ The answer is that you should not. It is just $\displaystyle \frac{6\left(\frac{2}{3}\right)^x+3}\frac{\left( \frac{2}{ 3}\right)^x+1}$
What do you mean i can't understand?
Originally Posted by srirahulan What do you mean i can't understand? What do you mean? That is just basic algebra. Divide by $\displaystyle 3^x$. There is no reason to try limits if one can't do algebra.
Thank you.........
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