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Question: How do you find the limit of \( {e^{\dfrac{1}{x}}} \) as x approaches \( {0^ - } \) ?...

How do you find the limit of e1x{e^{\dfrac{1}{x}}} as x approaches 0{0^ - } ?

Explanation

Solution

Here in this question we need to find the limit for the function. Let we define the given function as f(x)f(x) and we apply the limit to it. The function will get closer and closer to some number. When the x approaches to 0{0^ - } we need to find the limit of a function.

Complete step by step explanation:
The idea of a limit is a basis of all calculus. The limit of a function is defined as let f(x)f(x) be a function defined on an interval that contains x=ax = a . Then we say that limxaf(x)=L\mathop {\lim }\limits_{x \to a} f(x) = L , if for every ε>0\varepsilon > 0 there is some number
δ>0\delta > 0 such that f(x)L<ε\left| {f(x) - L} \right| < \varepsilon whenever 0<xa<δ0 < \left| {x - a} \right| < \delta .
Here we have to find the value of the limit when x is zero. Let we define the given function as f(x)=e1xf(x) = {e^{\dfrac{1}{x}}} . Now we are going to apply the limit to the function f(x)f(x) so we have
limx0f(x)=limx0e1x\mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {e^{\dfrac{1}{x}}}
The function f(x)f(x) is an exponential function. As x approaches to 0{0^ - } we have to substitute the x
as 0{0^ - } .
When we consider the x value we will consider it as a negative value because they have given 0{0^ - }
limx0f(x)=e10\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = {e^{ - \dfrac{1}{0}}}
Any number divided by zero then it becomes infinity, then
limx0f(x)=e\Rightarrow \mathop {\lim }\limits_{x \to {0^ - }} f(x) = {e^{ - \infty }}
By the exponential property we have e=0{e^{ - \infty }} = 0 . By considering this property we have
limx0f(x)=0\mathop {\lim }\limits_{x \to {0^ - }} f(x) = 0
Therefore we have found the limit for the function f(x)=e1xf(x) = {e^{\dfrac{1}{x}}}

Hence limx0e1x=0\mathop {\lim }\limits_{x \to {0^ - }} {e^{\dfrac{1}{x}}} = 0

Note: The plus and minus sign refer to the direction from which the function approaches zero. If they mention 0{0^ - } the limit as x approaches 0 from the negative side or from below. If they mention 0\+{0^ \+ } the limit as x approaches 0 from the positive side or from above. The property of exponential function should be known to solve this problem.