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Evaluate the following limit : \(\lim\limits_{\text x \to0}\cfrac{a^{m\text x}-b^{n\text x}}{sin\,k\text x} \)

lim(x→0) (amx - bnx)/(sin kx)

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As we need to find \(\lim\limits_{\text x \to0}\cfrac{a^{m\text x}-b^{n\text x}}{sin\,k\text x} \)

lim(x→0) (amx - bnx)/(sin kx)

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 = \(\lim\limits_{\text x \to0}\cfrac{a^{m\text x}-b^{n\text x}}{sin\,k\text x} \)

\(=\cfrac00\)(indeterminate form)

∴ 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 and logarithmic limits and also use of sandwich theorem - \(\lim\limits_{\text x \to 0}\cfrac{sin\,\text x}{\text x}=1\)

To get the desired forms, we need to include mx and nx as follows:

As Z = \(\lim\limits_{\text x \to0}\cfrac{a^{m\text x}-b^{n\text x}}{sin\,k\text x} \)

\(\Rightarrow\) Z 

{Adding and subtracting 1 in numerator}

{using algebra of limits}

To get the form as present in the formula we multiply and divide x into both terms respectively:

Use the formula:

Hence,

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