# For `f(t)=4t^2+3t+1` , find `lim_(h->0)(f(t+h)-f(t))/h` and for `f(x)=1/sqrt(5x+1)` , find `lim_(h->0)(f(3+h)-f(3))/h`

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### 1 Answer

For `f(t)=4t^2+3t+1`

`lim_(h->0)(f(t+h)-f(t))/h`

=> `lim_(h->0)(4(t+h)^2+3(t+h)+1 - 4t^2-3t-1)/h`

=> `lim_(h->0)(4(t+h-t)(t+h+t) + 3(t + h - t) + 1 - 1)/h`

=> `lim_(h->0)(4(h)(2t+h) + 3(h))/h`

=> `lim_(h->0) 4(2t+h) + 3`

=> `8t + 3`

**For `f(t)=4t^2+3t+1` , `lim_(h->0)(f(t+h)-f(t))/h = 8t + 3` **

For `f(x)=1/sqrt(5x+1)` ,

`lim_(h->0)(f(3+h)-f(3))/h`

`lim_(h->0)(1/sqrt(5(3+h)+1) - 1/sqrt(15+1))/h`

`lim_(h->0)(1/sqrt(16 + 5h) - 1/sqrt 16)/h`

`lim_(h->0)((sqrt 16 - sqrt(16 + 5h))/(sqrt(16+5h)*sqrt 16))/h`

substitute h = 0

=> `(sqrt 16 - sqrt 16)/(sqrt 16*sqrt 16)`

=> 0

**The value of ` lim_(h->0)(f(3+h)-f(3))/h = 0` **