Question

Find the exact value of the following limit: lim x approaches 0 e^6x... check photo above.​

Answer 1

Substituting x = 0 directly gives the indeterminate form 0/0, so you can use l'Hopital's rule (and you'll have to do that twice):

`\displaystyle\lim_(x\to0)(e^(6x)-6x-1)/(x^2)=\lim_(x\to0)(6e^(6x)-6)/(2x)=\lim_(x\to0)\frac{36e^(6x)}2`

The limand is continuous everywhere, so you can plug in x = 0 to get 36•1/2 = 18.