Question to test your Finding Limits Algebraically knowledge

Question

The hyperbolic sine function, sinh⁡(x)\sinh\left(x\right), models the velocity of a particle in hyperbolic motion. It is defined as sinh⁡(x)=ex−e−x2\sinh\left(x\right)=\frac{$$e^{x}$-e^{-x}$$}{2}. Find the following limits:lim⁡x→∞sinh⁡(x){\displaystyle\lim_{x\rightarrow\infty}\sinh\left(x\right){}}lim⁡x→−∞sinh⁡(x){\displaystyle\lim_{x\rightarrow-\infty}\sinh\left(x\right){}}

Accepted Answer

lim⁡x→∞sinh⁡(x){\displaystyle\lim_{x\rightarrow\infty}\sinh\left(x\right){}} =∞=\inftylim⁡x→−∞sinh⁡(x){\displaystyle\lim_{x\rightarrow-\infty}\sinh\left(x\right){}} =−∞=-\infty

Answer

lim⁡x→∞sinh⁡(x){\displaystyle\lim_{x\rightarrow\infty}\sinh\left(x\right){}} =0=0 lim⁡x→−∞sinh⁡(x){\displaystyle\lim_{x\rightarrow-\infty}\sinh\left(x\right){}} =0=0

Answer

lim⁡x→∞sinh⁡(x){\displaystyle\lim_{x\rightarrow\infty}\sinh\left(x\right){}} =1=1lim⁡x→−∞sinh⁡(x){\displaystyle\lim_{x\rightarrow-\infty}\sinh\left(x\right){}} =−1=-1

Answer

lim⁡x→∞sinh⁡(x){\displaystyle\lim_{x\rightarrow\infty}\sinh\left(x\right){}} =−1=-1lim⁡x→−∞sinh⁡(x){\displaystyle\lim_{x\rightarrow-\infty}\sinh\left(x\right){}} =1=1

Finding Limits Algebraically: Videos & Practice Problems