Question

Given $\dfrac{\left( {{e}^{h}} \right)-1}{h}$ how do you find the limit as $h$ approaches $0$ ?

Answer 1

Hint: Since, after applying the limit in the given function, we will get value in the form of $\dfrac{0}{0}$ . So, for getting the limit value of the given equation, first of all we will use the expansion of ${{e}^{h}}$ . Then, we will simplify the given function with use of expansion. After that we will use the limit and will get the required value of limit $h$ approaches $0$ .

Complete step-by-step solution:
Since, we have the given function as:
$\Rightarrow \dfrac{\left( {{e}^{h}} \right)-1}{h}$
If we apply the limit here, we will get the limit value in the form of $\dfrac{0}{0}$ . So, we will expand the ${{e}^{h}}$ as:
$\Rightarrow {{e}^{h}}=1+h+\dfrac{1}{2!}{{h}^{2}}+\dfrac{1}{3!}{{h}^{3}}+...$
Since, we already have the expansion of ${{e}^{h}}$ , we will use this value into the given function to make the solution easy as:
$\Rightarrow \dfrac{1+h+\dfrac{1}{2!}{{h}^{2}}+\dfrac{1}{3!}{{h}^{3}}+...-1}{h}$
Here, we can see that $-1$ will cancel out $1$ . So, we will have the above step below as:
$\Rightarrow \dfrac{h+\dfrac{1}{2!}{{h}^{2}}+\dfrac{1}{3!}{{h}^{3}}+...}{h}$
In the numerator, we can take a common factor that is $h$ because it is available in every term of the numerator as:
$\Rightarrow \dfrac{h\left( 1+\dfrac{1}{2!}h+\dfrac{1}{3!}{{h}^{2}}+... \right)}{h}$
Now, here we can see that the numerator and denominator have the same variable that is $h$ . So, we can cancel out it as:
$\Rightarrow 1+\dfrac{1}{2!}h+\dfrac{1}{3!}{{h}^{2}}+...$
Here, we got the simplified value of the given function. Therefore, we will apply limit as $h$ approaches $0$ now as:
$\Rightarrow \displaystyle \lim_{h \to 0}\left( 1+\dfrac{1}{2!}h+\dfrac{1}{3!}{{h}^{2}}+... \right)$
Now, we will use the limit and get the value as $1$ since all the terms multiple with $h$ will be zero because the limit is $h$ approaches $0$ as:
$\Rightarrow 1+\dfrac{1}{2!}\times 0+\dfrac{1}{3!}\times 0+...$
As we know that, excluding $1$ , all the terms are multiple of $h$, will give value zero as above. Thus, we have the final value of limit as:
$\Rightarrow 1$
Hence, the limit as $h$ approaches $0$in the given function $\dfrac{\left( {{e}^{h}} \right)-1}{h}$ will be $1$.

Note: In this type of question when we get the value of the limit for given function in the form of $\dfrac{0}{0}$ , we will use the another method of solving this type of question that are take factorization of given function, use conjugate of numerator or denominator, L-Hospital Rule, etc.