Exponential Function sketching (1 Viewer)

[Hello_World2]

for question 8 as x--> ∞ why does y approach 0 and not 1 ? cus if i let x=∞ then I will get ∞/∞ which equals 1 ?

 

[Hamburgler]

As x-> inf, y approaches inf as e^x is far bigger than x in e^x/x so numerator overpowers it significantly
As x-> -inf, e^x approaches 0, hence y approaches 0 from the bottom

 

[Hello_World2]

Ohh, so does it depend on the case and this will be different for other functions ?

 

[yanujw]

Infinity/Infinity is not the appropriate way to find limits, you need to consider dominance laws.

Exponentials dominate algebraic terms which dominates logarithms. And the dominant factor is the one that dictates how the function behaves as x approaches infinity.

 

[cossine]

View attachment 36574
for question 8 as x--> ∞ why does y approach 0 and not 1 ? cus if i let x=∞ then I will get ∞/∞ which equals 1 ?

A lot of operations involving infinity are undefined.

infinity + infinity = infinity (correct?)

Then infinity/ infinity = (infinity + infinity )/infinity = 2

For this reason division by infinity is undefined.

Any limit that cannot be evaluated upon substitution is considered indeterminant. We exclude case like lim x->0 1/x^2 as these can easily be evaluated using the definition of infinite limits.

So the question arises how to evaluate indeterminant limits. The answer comes to knowing theorems like lim x->0 sin(x)/x = 1 and techniques like rationalisation or L'Hopital rule. So one way to approach the question is first check L'hopital rule can be used and the apply.