Question

How do you find the limit of $ {x^{2x}} $ as x approaches 0?

Answer 1

Hint: In order to determine the limit of the above function, Consider \[y = {x^{2x}}\], and take natural log on both sides. Now you can see by putting $ x = 0 $ the result is in the indeterminate form i.e. $ \dfrac{0}{0}\,or\,\dfrac{{ \pm \infty }}{{ \pm \infty }} $ . To remove the indeterminate form use the L-hospital’s rule which says $ \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}} $ . In our case, consider $ f(x) = \ln x $ and $ g(x) = \dfrac{1}{x} $
and calculate their derivative with the help of derivative rules $ \dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x} $ and $ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} $ put them into the L-hospital’s rule to obtain the limit of the function.

Complete step-by-step answer:
We are given an exponential function in variable $ x $ i.e. $ {x^{2x}} $ having limit $ x \to 0 $ .
 $ L = \mathop {\lim }\limits_{x \to 0} {x^{2x}} $
Consider \[y = {x^{2x}}\]so that if we take natural log on both of the sides, we get
\[\ln y = 2x\ln x\]
 $
\Rightarrow \ln L = \mathop {\lim }\limits_{x \to 0} \,\,2x\ln x \\
   = 2\mathop {\lim }\limits_{x \to 0} \,\,\dfrac{{\ln x}}{{\dfrac{1}{x}}} \\
  $
As you can see this is the limit problem, so if we directly put the limit $ x = 0 $ the result will have denominator and numerator both equal to zero i.e. of the indeterminate form $ \dfrac{{ \pm \infty }}{{ \pm \infty }} $
So to avoid the indeterminate form, we will be using L-Hospital’s rule
According to L-Hospital’s rule if any limit has the following cases
 $ \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{0}{0}\,\,\,\,\,\,or\,\,\,\,\,\,\mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \dfrac{{ \pm \infty }}{{ \pm \infty }} $ where a is any real number.
So in such cases we calculate limit as $ \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}} $
In our case, we have $ f(x) = \ln x $ and $ g(x) = \dfrac{1}{x} $
Now calculating the derivative of $ f\left( x \right) $ using the rule of derivative $ \dfrac{d}{{dx}}\left( {\ln x} \right) = \dfrac{1}{x} $
 $ f'\left( x \right) = \dfrac{1}{x} $ and
Derivative of $ g\left( x \right) $ by using rule $ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} $ , we get
 $ \Rightarrow g'\left( x \right) = - \dfrac{1}{{{x^2}}} $
Putting these $ f'\left( x \right) $ and $ g'\left( x \right) $ in the L-Hospital rule , we get
 $ \ln L = 2\mathop {\lim }\limits_{x \to 0} \,\,\dfrac{{\ln x}}{{\dfrac{1}{x}}} = 2\mathop {\lim }\limits_{x \to 0} \,\,\dfrac{{\dfrac{1}{x}}}{{ - \dfrac{1}{{{x^2}}}}} $
Simplifying further , we get
 $ \Rightarrow \ln L = 2\mathop {\lim }\limits_{x \to 0} \,\,x = 0 $
So,
\[\Rightarrow \mathop {\lim }\limits_{x \to 0} \,\,\ln L = 0 \Rightarrow \mathop {\lim }\limits_{x \to 0} \,L = 1\]
Therefore, the limit of function $ {x^{2x}} $ as $ x \to 0 $ is equal to 1.
So, the correct answer is “1”.

Note: Limit: You and your companions choose to meet at some spot outside. Is it essential that every one of your companions is living in a similar spot and stroll on a similar street to arrive at that place? Actually no, not generally. All companions come from various pieces of the city or nation to meet at that one single spot.
It would appear that intermingling of various components to a solitary point. Mathematically, it resembles an intermingling of a function to a specific value. It is an illustration of cutoff points. Cut-off points show how a few functions are limited. The function watches out for some worth when its breaking point moves toward some value.
1. L-Hospital’s Rule:
 $ \mathop {\lim }\limits_{x \to a} \dfrac{{f(x)}}{{g(x)}} = \mathop {\lim }\limits_{x \to a} \dfrac{{f'(x)}}{{g'(x)}} $
2. $ \dfrac{d}{{dx}}\left( x \right) = 1 $
3. $ \dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x} $
4. $ \dfrac{d}{{dx}}\left( c \right) = 0 $