Evaluate the following limit : lim(x→0) (e^3 + x - sin x - e^3)/x

Question

Evaluate the following limit : \(\lim\limits_{\text x \to0}\cfrac{e^{3+\text x}-sin\,\text x - e^3}{\text x} \)

lim(x→0) (e^{3 + x} - sin x - e³)/x

Answer 1

As we need to find \(\lim\limits_{\text x \to0}\cfrac{e^{3+\text x}-sin\,\text x - e^3}{\text x} \)

lim(x→0) (e^{3 + x} - sin x - e³)/x

We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞, .. etc.)

Let Z =  \(\lim\limits_{\text x \to0}\cfrac{e^{3+\text x}-sin\,\text x - e^3}{\text x} \)\(=\cfrac{e^{3+0}-sin\,0-e^3}0\) 

 \(=\cfrac00\)(indeterminate)

∴ we need to take steps to remove this form so that we can get a finite value.

TIP: Most of the problems of logarithmic and exponential limits are solved using the formula

   \(\lim\limits_{\text x \to0}\cfrac{a^{\text x-1}}{\text x}\) = log a and \(\lim\limits_{\text x \to0}\cfrac{log(1+\text x)}{\text x}=1\)

This question is a direct application of limits formula of exponential limits.

As Z = \(\lim\limits_{\text x \to0}\cfrac{e^{3+\text x}-sin\,\text x - e^3}{\text x} \) 

{using algebra of limits}

Use the formula:   \(\lim\limits_{\text x \to0}\cfrac{a^{\text x-1}}{\text x}\) = log a and \(\lim\limits_{\text x \to0} \cfrac{sin\,\text x}{\text x}=1\)

∴ Z = e³ log e – 1

{∵ log e = 1}

Hence,

\(\lim\limits_{\text x \to0}\cfrac{e^{3+\text x}-sin\,\text x - e^3}{\text x} \) = e³ - 1