Calculus Limit

Question

Find the value of

 

Here I got till $lim_{x \rightarrow \infty} \frac{4^2 + \frac{3^x}{4^x}}{4^{-2}}$. But how to proceed further?

Comments

@Shubhanshu

u already done everything

just put (3/4)^x=0

and evaluate the result

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if your query answered than plz share screenshot here too

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Thanks @srestha what Idea you gave is nice, But I want to know that what is the derivation of $\lim_{x\rightarrow \infty}(\frac{3}{4})^x$.

Even I have tried. when x $\rightarrow \infty$ then t $\rightarrow 0$ method, and also did using log, but not getting equal to 0.

Answer 1

$\:a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)}$

= $\left(\frac{3}{4}\right)^x=e^{x\ln \left(\frac{3}{4}\right)}$

$=\lim _{x\to \infty \:}\left(e^{x\ln \left(\frac{3}{4}\right)}\right)$

$=\lim _{x\to \infty \:}\left(e^{x\ln \left(\frac{4}{3}\right)^{-1}}\right)$

$=\lim _{x\to \infty \:}\left(e^{-x\ln \left(\frac{4}{3}\right)}\right)$

=  $e^{-\infty }$ 

Take  x towards  $-\infty$  ,$e^{-\infty }$  approaches zero .@your line  and also did using log, but not getting equal to 0 . No chance that you will get $\left ( \frac{3}{4} \right )^{x} =0$.            Hence proved  Thank you !!

Comments

Thanks !!!!