Question

Evaluate : \[{\lim _{x \to 0}}{\left( {\dfrac{{tanx}}{x}} \right)^{\dfrac{{tanx}}{{tanx - x}}}}\].

Answer 1

Hint: To solve this kind of question, first we need to know the type of the indeterminate form of the limit. Then only we move ahead. After that try to multiply and divide the exponential part by \[x\] and add and subtract the \[\left( {\dfrac{{tanx}}{x}} \right)\] part with \[1\]. Then use some general properties and formulas of limit, to reach the desired result.

Complete step-by-step solution:
First we need to know the type of the indeterminate form of the limit. So we put the value \[x = 0\] in \[{\lim _{x \to 0}}{\left( {\dfrac{{tanx}}{x}} \right)^{\dfrac{{tanx}}{{tanx - x}}}}\]. We find out that this is \[\dfrac{0}{0}\] form. Now we try to make this into a deterministic form. For that we multiply and divide the exponential part by \[x\] and add and subtract the \[\left( {\dfrac{{tanx}}{x}} \right)\] part with \[1\], shown below,
\[
  {\lim _{x \to 0}}{\left( {1 + \dfrac{{tanx}}{x} - 1} \right)^{\left( {\dfrac{x}{{tanx - x}}} \right)\left( {\dfrac{{tanx}}{x}} \right)}} \\
   \Rightarrow {\lim _{x \to 0}}{\left( {1 + \dfrac{{\tan x - x}}{x}} \right)^{\left( {\dfrac{x}{{tanx - x}}} \right)\left( {\dfrac{{tanx}}{x}} \right)}} \\
 \]
Now, when we see above step carefully, we see that it is of the form \[{\lim _{x \to 0}}{\left( {1 + x} \right)^{\dfrac{1}{x}}}\], and we know that
\[{\lim _{x \to 0}}{\left( {1 + x} \right)^{\dfrac{1}{x}}} = e\].
So, using the above formula we move ahead as,
\[
   \Rightarrow {\lim _{x \to 0}}{e^{\left( {\dfrac{{tanx}}{x}} \right)}} \\
   \Rightarrow {e^{{{\lim }_{x \to 0}}\dfrac{{tanx}}{x}}} \\
 \]
Now, since we know that, \[{\lim _{x \to 0}}\dfrac{{\tan x}}{x} = 1\], so using this we get,
\[{\lim _{x \to 0}}{\left( {\dfrac{{tanx}}{x}} \right)^{\dfrac{{tanx}}{{tanx - x}}}} = e\]
So, the value of the expression asked in the question is \[e\].
Formula used: We have used the following written in the solution above
\[{\lim _{x \to 0}}\dfrac{{\tan x}}{x} = 1\]
\[{\lim _{x \to 0}}{\left( {1 + x} \right)^{\dfrac{1}{x}}} = e\]

Note: While solving the questions related to limits, if we are getting a deterministic form after directly putting the tending value of variable, we simply consider the result through that as the value of limit. Limit is the mathematical concept based on the idea of closeness which is used generally and preferably to assign values to a certain function at those points where no values are defined, in such a way as to be consistent with nearby values.