Limits of functions III

Calculus Level 2

Find the limit :

lim ⁡ x → + ∞ ( e x − e − x e x + e − x ) . \large \lim _{x\to+\infty}\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right).

Use this to the problem: lim ⁡ x → + ∞ e x . = + ∞ \lim\limits_{x\to +\infty}e^x.=+\infty


The answer is 1.

Dimitris Kouroupos by dimitris kouroupos , 23, Greece

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1 solution

THE LIMIT IS 0 0 \frac{0}{0}

  • next we developt the limit :

lim ⁡ x → + ∞ ( e x − e − x e x + e − x ) = lim ⁡ x → + ∞ ( e 2 x − 1 e 2 x + 1 ) = lim ⁡ x → + ∞ ( e 2 x e 2 x + 1 ) − lim ⁡ x → + ∞ ( 1 e 2 x + 1 ) = lim ⁡ x → + ∞ ( e 2 x e 2 x ( 1 + e − 2 x ) ) = lim ⁡ x → + ∞ ( 1 1 + e − 2 x ) = 1 \large \lim _{x\to+\infty }\left(\frac{e^x-e^{-x}}{e^x+e^{-x}}\right)= \large \lim _{x\to+\infty }\left(\frac{e^{2x}-1}{e^{2x}+1}\right)=\large \lim _{x\to+\infty }\left(\frac{e^{2x}}{e^{2x}+1}\right)-\large \lim _{x\to+\infty }\left(\frac{1}{e^{2x}+1}\right)=\large \lim _{x\to+\infty }\left(\frac{e^{2x}}{e^{2x}(1+e^{-2x})}\right)= \large \lim _{x\to+\infty }\left(\frac{1}{1+e^{-2x}}\right)=1

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