Question

Find the exact value of the following limit: lim x goes to 0 (e^8x-8x-1)/x^2

Answer 1

32

Explanation:

We can apple L'Hopital's Rule to solve this problem:

`\lim _(x\to \:\:0)\left((e^(8x)-8x-1)/(x^2)\right)\\=> \lim _(x\to \:0)\left((8e^(8x)-8)/(2x)\right)\\\\=> \lim _(x\to \:0)\left((4\left(e^(8x)-1\right))/(x)\right)\\\\=> \lim _(x\to \:0)\left((32e^(8x))/(1)\right)\\\\=>(32e^(8\cdot \:0))/(1)\\\\=> 32`

Hope this helps!