Question

Evaluate:
$\rm \displaystyle \lim_{x \to 0}\ \frac{3^{x}+5^{x}-2^{x+1}}{\sin x}$

Answer 1

we have to evaluate, $\rm \displaystyle \lim_{x \to 0}\ \frac{3^{x}+5^{x}-2^{x+1}}{\sin x}$

check form of the limit.

putting, x = 0 , in f(x) = {3^x + 5^x - 2^(x+1)}/sinx

f(0) = {3^0 + 5^0 - 2^(0 + 1)}/sin(0)

= {1 + 1- 2}/0

= 0/0 hence, it is in the form of limit.

now, $\rm \displaystyle \lim_{x \to 0}\ \frac{3^{x}+5^{x}-2^{x+1}}{\sin x}$

= $\displaystyle\lim_{x\to 0}\frac{(3^x+5^x-2.2^x}{sinx}$

= $\displaystyle\lim_{x\to 0}\frac{\frac{3^x-1}{x}\times x+\frac{5^x-1}{x}\times x-2\frac{2^x-1}{x}\times x}{x\frac{sinx}{x}}$

we know,

$\displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=loga$

and $\displaystyle\lim_{x\to 0}\frac{sinx}{x}=1$

= $\displaystyle\lim_{x\to 0}\frac{xlog3+xlog5-2xlog2}{x.1}$

= (log3+ log5 - 2log2)/1

= {log(3 × 5) - log2²}

= log(15/4) [ans]

Answer 2

Answer:

log($\frac{15}{4}$)

Step-by-step explanation:

In this question,

We have to evaluate, $\rm \displaystyle \lim_{x \to 0}\ \frac{3^{x}+5^{x}-2^{x+1}}{\sin x}$

Now we will check form of the limit.

Putting, x = 0 , in f(x) = $\frac{3^{x} + 5^{x} - 2^{(x+1)}}{sinx}$

f(0) =$\frac{3^{0} + 5^{0} - 2^{(0 + 1)}}{sin(0)}$

= $\frac{{1 + 1- 2}}{0}$

= 0/0 hence, it is in the form of limit.

Now,  $\rm \displaystyle \lim_{x \to 0}\ \frac{3^{x}+5^{x}-2^{x+1}}{\sin x}$

= $\displaystyle\lim_{x\to 0}\frac{(3^x+5^x-2.2^x}{sinx}$

= $\displaystyle\lim_{x\to 0}\frac{\frac{3^x-1}{x}\times x+\frac{5^x-1}{x}\times x-2\frac{2^x-1}{x}\times x}{x\frac{sinx}{x}}$

We know,

$\mathbf{\displaystyle\lim_{x\to 0}\frac{a^x-1}{x}=loga}$

and $\mathbf{\displaystyle\lim_{x\to 0}\frac{sinx}{x}=1}$

= $\displaystyle\lim_{x\to 0}\frac{xlog3+xlog5-2xlog2}{x.1}$

= $\frac{(log3+ log5 - 2log2)}{1}$

= {log(3 × 5) - log2²}

= log($\frac{15}{4}$)