Use a numerical method to estimate the value of the limit (if it exists.) lim e^5t -1 /t as x tends to 0 from both sides

Hello from MrBillDoesMath!

5

Discussion:

Consider the expansion of e^x:

e^x = 1 + x + x^2/2 + x^3/6 +            =>      replace x by 5t

e^(5t) = 1 + (5t)  + (5t)^2/2 +              =>      subtract 1 from both sides

e^(5t) - 1 = (5t) + (5t)^2/2+                  =>      divide both sides by t

(e^(5t) -1)/ t = 5 + (25/2)  t +

so as t ends to 0  the quotient tends to    

5 + (25/2)0 + (other terms) *0   ->   5

Thank you,

MrB

fahrenheit

step-by-step explanation:

fahrenheit is more precise than celsius. the ambient temperature on most of the inhabited world ranges from -20 degrees fahrenheit to 110 degrees fahrenheit—a 130-degree range. on the celsius scale, that range is from -28.8 degrees to 43.3 degrees—a 72.1-degree range. this means that you can get a more exact measurement of the air temperature using fahrenheit because it uses almost twice the scale.

i hope this is what you were looking for.

see attached graph